22 worked examples31 exercises7 figuresbölüm 1 / 14
01Limits: definition, computation, one-sided and infinite limits
Drop a stone from a bridge deck fifty metres above the water. Between the 2nd and the 3rd second it falls exactly $24.5$ m, so over that second it averaged $24.5$ m/s — but a speed gun pointed at it at the single instant $t=2$ would not read $24.5$. Ask what it reads at that instant and the arithmetic collapses: $0$ metres travelled in $0$ seconds, which is not a number.
By the end of this section you can turn that $0/0$ into an exact reading — $19.6$ m/s — and defend every line of the computation the way a grader wants it defended.
In 60 seconds
A is the number the outputs of $f$ close in on while $x$ closes in on $a$: try substitution first, and when substitution is illegal, rewrite $f$ until it becomes legal.
$m=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$
speed at an instant, slope of a curve at one point
a bounded oscillating factor times something that goes to $0$
Three most common mistakes
Writing $0/0$ as the answer. $0/0$ is not a value, it is the instruction rewrite this expression.
Reporting $\lim_{x\to 0}\frac{1}{x}=\infty$. The left side runs to $-\infty$ and the right side to $+\infty$, so there is no single answer — only two one-sided ones.
Assuming $\lim_{x\to a}f(x)=f(a)$. The limit never looks at $f(a)$; the two agree only when $f$ is continuous at $a$, which is a separate claim you have to check.
Grade weights this term: Midterm 1 28%, Midterm 2 28%, Final 28%, quizzes 10%, homework 6%.
How much time do you have?
10 minutes
The first line of every limit question: substitute, read the form that comes out, and apply the one rewrite that clears most of them.
card, c-limit-laws, formula card
45 minutes
Every type that can be set: the two sided existence test, the sign work at an asymptote, the three part continuity check, and one full ladder from a worked solution to a bare one.
card, c-limit-laws, c-one-sided, c-infinite-limits, c-continuity, faded ladder, formula card
full read
Where the limit came from, why each law is allowed, and enough mixed practice that you pick the method yourself instead of being told it.
hook, prerequisites, c-instantaneous-rate, c-limit-idea, c-one-sided, c-infinite-limits, c-limit-laws, c-squeeze, c-continuity, contrast pairs, faded ladder, exam example, practice A to D
By the end of this section
Convert an average rate over an interval into an instantaneous rate at a point, and say exactly why the one point computation returns $0/0$.
Read a limit off a table or a graph and state what the limit ignores about $f(a)$.
Compute of piecewise and absolute value functions and apply the existence test.
Locate and give the correct sign of every one sided .
Justify each line of a limit computation by naming the limit law or the algebraic rewrite that licenses it.
Bound an oscillating function between two functions with a common limit and finish with the Squeeze Theorem.
Test with the three part definition and use it to license direct substitution.
Syllabus coverage
1.4
The tangent and velocity problems
The falling ball and the slope of a curve at a single point are the whole content of this token, and they are the reason limits exist.
covered
1.5
The limit of a function: informal definition, one sided limits, infinite limits
Split over three blocks: the definition and what it ignores, then the one sided limits and the existence test, then infinite limits and vertical asymptotes.
covered
1.6
Calculating limits using the limit laws, and the Squeeze Theorem
The laws themselves, the $0/0$ repair kit built on top of them, and the Squeeze Theorem in its own block.
covered
1.8
Continuity
The part a limits section needs is here: the three part test at a point and the substitution property it licenses. Classifying discontinuities, continuity of compositions and the Intermediate Value Theorem carry on in the next section.
covered
1.7
The precise (epsilon-delta) definition of a limit
The weekly plan jumps from 1.6 to 1.8, so the epsilon-delta machinery is not something you are responsible for this term. Every computation in this course runs on the informal definition below.
off_syllabus
Recall first
Slope through two points
$m=\dfrac{y_2-y_1}{x_2-x_1}$, and it needs two distinct points.
Every secant slope in the opening problem is this formula with both points sitting on the curve.
The three factoring patterns that do most of the work
A $0/0$ quotient of polynomials always hides the factor $(x-a)$ in both parts; these patterns expose it.
of a root expression
$(\sqrt{u}-v)(\sqrt{u}+v)=u-v^2$ — the root disappears.
It is the only move that turns $\sqrt{x+4}-2$ into something with a visible factor of $x$.
Absolute value is a piecewise formula
$|t|=\begin{cases}t,& t\ge 0\\ -t,& t<0\end{cases}$, so $|x-a|=x-a$ when $x>a$ and $|x-a|=-(x-a)$ when $x
Every absolute value in this section is unpacked with this before anything is cancelled.
Sine and cosine are trapped between $-1$ and $1$
$-1\le \sin t\le 1$ and $-1\le \cos t\le 1$ for every real $t$, no matter how wild $t$ is.
The Squeeze Theorem needs a bound to start from, and for oscillating factors this is the bound.
Sign of a quotient from the signs of its parts
A fixed nonzero number over a very small positive number is very large positive; over a very small negative number it is very large negative.
This is all the sign reasoning an infinite limit ever needs.
Try it yourself first (3 questions)
1§01.2 — the value at a point versus the value approached●●○○○
A first year student simplifies $f(x)=\dfrac{x^2-4}{x-2}$ to $x+2$ on the board and then writes $f(2)=4$ underneath. Something in that pair of statements is wrong, and it is worth finding out which part before we build anything on it.
Given
$f(x)=\dfrac{x^2-4}{x-2}$, with no extra definition at $x=2$
Find
(a) What are $f(2)$ and $\lim_{x\to 2}f(x)$?
Hint 1/4
Two separate questions are hiding here: what does the formula produce at $x=2$, and what do the outputs do near $x=2$. Answer them one at a time.
Hint 2/4
Substituting $x=a$ into a formula is legal only when the formula is defined there. Cancelling a factor $(x-2)$ changes the domain, so the cancelled version is a different function.
Hint 3/4
Put $x=2$ into the original: numerator $2^2-4=0$, denominator $2-2=0$. Now try neighbours: $f(1.9)=3.9$, $f(1.99)=3.99$, $f(2.01)=4.01$.
Hint 4/4
So $f(2)$ is undefined while the nearby outputs close in on $4$.
Show solutionTest the formula at the point itself
$f(2)=\frac{2^2-4}{2-2}=\frac{0}{0}$
division by zero is undefined, so $2$ is simply not in the domain of $f$
Ask what happens next to the point
$\frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2$
legal for every $x\neq 2$, and $x\neq 2$ is exactly the regime a limit looks at
$\lim_{x\to 2}(x+2)=4$
the cancelled expression is a polynomial, so its nearby values close in on its value at 2
Neighbours confirm it from both sides: $f(1.99)=3.99$ and $f(2.01)=4.01$, and neither side ever reaches 4.
2§01.4 — what the symbol infinity is allowed to mean●●○○○
Calculators plot $y=1/x$ as a curve that shoots up on the right of the origin and plunges on the left. A very common exam answer for the behaviour at $0$ is a single symbol. Pick the honest description.
Given
$f(x)=\dfrac{1}{x}$, and $x$ approaching $0$
Find
(a) What is $\lim_{x\to 0}\dfrac{1}{x}$?
Hint 1/4
The two sides of $0$ are different neighbourhoods. Check them separately before answering as if they were one.
Hint 2/4
A two sided limit is allowed to exist only when both one sided behaviours agree.
Hint 3/4
Right side: $f(0.01)=100$, $f(0.001)=1000$, growing without bound. Left side: $f(-0.01)=-100$, $f(-0.001)=-1000$, falling without bound.
Hint 4/4
The sides disagree, so there is no single answer: the limit does not exist, and the honest report is $\lim_{x\to0^{-}}=-\infty$, $\lim_{x\to0^{+}}=+\infty$.
Show solutionTake the sides apart
$\lim_{x\to 0^{+}}\frac1x=+\infty$
the numerator is fixed at 1 and the denominator is a shrinking positive number
$\lim_{x\to 0^{-}}\frac1x=-\infty$
same size, opposite sign: the denominator is a shrinking negative number
Apply the existence test
$+\infty\neq-\infty$
the two one sided behaviours do not agree on anything, not even on a direction
Answer $$\boxed{\lim_{x\to 0}\frac1x\ \text{does not exist}}$$
Check
A sign check settles it without any calculation: $1/x$ is an odd function, so whatever it does on one side of $0$ it does with the opposite sign on the other.
3§01.1 — what an average rate does and does not tell you●○○○○
A drone flies down a straight corridor. Its logger records only two rows: position $0$ m at $t=0$ s and position $32$ m at $t=4$ s. Someone reads those two rows and announces the drone's speed.
Given
position 0 m at t = 0 s
position 32 m at t = 4 s
nothing recorded in between
Find
(a) Which single statement is guaranteed by these two rows?
Hint 1/4
Ask what a two row table can possibly measure: something about the whole interval, or something about one instant inside it?
Hint 2/4
Average velocity over $[t_1,t_2]$ is $\dfrac{s(t_2)-s(t_1)}{t_2-t_1}$, and every quantity computed from two rows is a statement about the interval between them.
Hint 3/4
With the given rows: $\dfrac{32-0}{4-0}=8$ m/s, an interval statement about $[0,4]$.
Hint 4/4
So the only guaranteed claim is that the average over the four seconds was 8 m/s.
$\text{plan B: hover for 3 s, then cover the 32 m in the last second}$
a completely different flight producing exactly the same two logged rows
$\text{during the first 3 s the speed is } 0\ \text{m/s}\neq 8\ \text{m/s}$
so the two rows cannot possibly decide the speed at an interior instant
Answer $$\boxed{\text{Only the average over }[0,4]\text{ is 8 m/s}}$$
Check
A third itinerary with the same two rows: cover all 32 m in the first second, then stop. The average over $[0,4]$ is still $8$ m/s, and at $t=4$ the drone is not moving at all.
This gap between over an interval and at an instant is the entire reason the next block exists.
Notation
symbol
reads as
means
watch out
$\lim_{x\to a}f(x)=L$
the limit of f of x as x approaches a equals L
the outputs close in on the number L while x closes in on a
It says nothing at all about $f(a)$, and it does not claim that $f(a)$ exists.
$x\to a^{-}$
x approaches a from the left
x is smaller than a and moving towards it
The minus marks a side, not a sign: $x\to 3^{-}$ means values like $2.99$, not $-3$.
$x\to a^{+}$
x approaches a from the right
x is bigger than a and moving towards it
On a piecewise formula this side picks the branch valid for $x>a$, wherever the equality case is parked.
$\lim_{x\to a}f(x)=\infty$
f of x approaches infinity as x approaches a
the outputs pass every bound you name
$\infty$ is not a number, so this is a description of how the limit fails, not a value you may do arithmetic with.
$DNE$
does not exist
no single number describes the approach
An infinite limit is a DNE case too — writing $\infty$ just says more about the failure.
$\vert x\vert$
the absolute value of x
the distance from x to 0
Never cancel it against $x$: it is a piecewise formula, so unpack it before cancelling.
$\tfrac{0}{0}$
zero over zero, an
substitution failed and the expression must be rewritten
It is a diagnosis, never an answer, and it does not mean the limit fails to exist.
$f(a)$ vs $\lim_{x\to a}f(x)$
the value at a, versus the value approached at a
two independent numbers
They coincide exactly when $f$ is continuous at $a$; assuming it in advance is the most expensive habit in this chapter.
Conventions used here
Classifying a break at a point that is not in the domain.
When $a$ is not in the domain of $f$ we still classify the break at $a$ — removable, jump or infinite — and that is what a question asking you to classify the discontinuities wants. The sentence $f$ is continuous at every point of its domain stays true at the same time, and it is the sentence to write when the question asks whether $f$ is continuous. So $1/x$ has an infinite discontinuity at $0$ and is also continuous on its domain; both are correct, and they answer different questions.
Two readings of the word discontinuous are in circulation. Mixing them inside one script is what attaches a wrong sentence to a correct calculation, so this page classifies breaks everywhere and never uses an undefined point as evidence that a function is badly behaved.
Reporting an infinite limit rather than DNE.
An infinite limit is one particular way of failing to exist. When both sides run off the same way we write $+\infty$ or $-\infty$, because that says more than DNE does. When the two sides run to opposite infinities we write DNE and then give the two one sided answers separately.
Both conventions are standard, and switching between them inside one question is what loses the marks that the sign discussion carries.
Near a excludes a; open and closed intervals.
The phrase for all $x$ near $a$ means: for all $x$ in some open interval around $a$, with the single point $a$ itself removed. Open intervals $(c,d)$ leave their endpoints out, closed intervals $[c,d]$ include them.
Every cancellation on this page is legal only because the point $a$ is excluded, so the exclusion is part of the rule and not a detail.
Radians, and the falling body model.
Every angle is measured in radians unless a degree sign is printed. The falling body used in the examples is $s(t)=4.9t^{2}$: metres fallen $t$ seconds after release from rest, with $g=9.8$ m/s$^2$ and air resistance ignored, so every rate read off it is in metres per second.
The verification step in those examples is a units and magnitude check, which only works if the model is fixed once and not changed half way through.
From a rate over an interval to a rate at an instant
What we have so far: the slope of a line through two points, and the average speed over a time interval. Both need two moments, and the question in front of us has only one.
Solvable with what we have
Average speed over $[2,3]$: $\dfrac{44.1-19.6}{3-2}=24.5$ m/s.
Average speed over $[2,2.1]$: $\dfrac{21.609-19.6}{0.1}=20.09$ m/s.
The slope through any two points of $s=4.9t^{2}$.
Not solvable yet
The speed at the single instant $t=2$.
The slope of $s=4.9t^{2}$ at the single point $(2,19.6)$.
Anything at all once the two points collapse into one.
Shrink the interval until it is no interval at all — both endpoints at $t=2$:
The short intervals gave $24.5$, then $20.09$, then $19.649$. The instant itself gives nothing.
Why it fails
The formula needs two distinct points; collapsing them makes distance and elapsed time both exactly $0$, and $0/0$ is no number — every $c$ satisfies $c\cdot 0=0$. So we stop trying to reach the instant and watch where the numbers head as we approach it.
DefinitionDefinition 1.1: Instantaneous rate of change and the slope of a curve
Conditions
$f$ is defined on an open interval around $a$
the quotient is formed only for $h\neq 0$ (equivalently $x\neq a$)
the numbers it produces close in on one value as the second point slides in
Take the average rate between $a$ and a nearby point, then ask what number those averages close in on as the nearby point slides into $a$. That number, if it exists, is the rate at $a$.
As the second point slides down towards P the secant slopes fall 24.5, 22.05, 19.6 — the tangent is what they close in on, not one of them.
Looks like this, but is not
This is a tangent: the line $y=2x-1$ meets $y=x^{2}$ at $(1,1)$ and has exactly the slope the secants close in on.
This looks like a definition of tangent but is not:the line touching the curve at exactly one point. The vertical line $x=1$ meets $y=x^{2}$ once and is tangent to nothing; the tangent to $y=x^{3}$ at the origin is $y=0$, which cuts through the curve. Counting intersections decides nothing.
h (s)
interval
average speed = 19.6 + 4.9h
1
[2, 3]
24.5
0.5
[2, 2.5]
22.05
0.1
[2, 2.1]
20.09
0.01
[2, 2.01]
19.649
-0.01
[1.99, 2]
19.551
-0.1
[1.9, 2]
19.11
From above the numbers fall towards 19.6, from below they rise towards it, and neither side ever produces it. It is not a row of the table; it is what the rows are aiming at.
Speed of the dropped ball at exactly t = 2 s
A ball is dropped and falls $s(t)=4.9t^{2}$ metres in $t$ seconds. Find its speed at the instant $t=2$.
Given
$s(t)=4.9t^{2}$ (metres, seconds)
the instant $a=2$
Find
The instantaneous speed at t = 2, in m/s.
SolutionBuild the average speed over a step of size h
expanding first is what makes the $19.6$ cancel; without expanding, the $h$ in the denominator has nothing to cancel against
$=19.6+4.9h\qquad (h\neq 0)$
cancelling $h$ is legal because $h\neq 0$ throughout — and that restriction is exactly the regime a limit looks at
Let the step shrink
$\lim_{h\to 0}\left(19.6+4.9h\right)=19.6$
the repaired expression is a polynomial in $h$, so the nearby values close in on its value at $h=0$
Answer $$\boxed{v(2)=19.6\ \text{m/s}}$$
Check
Compare with the table: from above 19.649, from below 19.551, and 19.6 sits between every such pair — the two sides trap it.
The algebra was three lines, and all three existed for one purpose: to make the h in the denominator cancel before the step is allowed to vanish.
This closes the question the section opened with. The average over the second was 24.5 m/s, the radar reading at the instant is 19.6 m/s, and the gap between them is the whole reason limits were invented.
Slope of the curve y = 1/x at the point (2, 0.5)
Find the slope of $y=\dfrac{1}{x}$ at $x=2$.
Given
$f(x)=1/x$
$a=2$
Find
The slope of the at x = 2.
SolutionForm the difference quotient and clear the small fractions
dividing by h is the same as multiplying by 1/h, and now the cancellation is available
Shrink the step
$\lim_{h\to 0}\frac{-1}{2(2+h)}=\frac{-1}{4}$
substitution is legal in the repaired expression: its denominator tends to $4\neq 0$
Answer $$\boxed{m=-\tfrac14}$$
Check
Sign and size both check out: $1/x$ is decreasing, so a negative slope was the only possibility, and the same computation at $x=1$ gives $-1$, a steeper fall — which is what the graph shows.
Checkpoint
§01.1 — average rate over a shrinking interval●○○○○
Stay with the falling ball for thirty seconds. We showed that the average speed over the interval from 2 to 2 + h is exactly 19.6 + 4.9h m/s, with no approximation anywhere in that formula.
Given
average speed over $[2,2+h]$ equals $19.6+4.9h$
$h=0.5$
Find
(a) What is the average speed over $[2,2.5]$, and is it larger or smaller than the instantaneous 19.6 m/s? One line for why.
Hint 1/4
You are not being asked for a limit here — only for one value of a formula you already have.
Hint 2/4
The average speed over $[2,2+h]$ is $19.6+4.9h$.
Hint 3/4
Here the step is $h=0.5$, so substitute it into $19.6+4.9h$.
Hint 4/4
The average is $22.05$ m/s, larger than $19.6$ m/s.
Show solutionSubstitute
$19.6+4.9(0.5)=19.6+2.45=22.05$
the formula is exact, so no approximation enters
$22.05>19.6$
the extra 2.45 m/s is the acceleration collected over the half second
⚠ Cancelling h before it is allowed, or setting h = 0 too early
the h in the denominator is annoying and the temptation is to remove it by substituting rather than by algebra
wrong$\frac{s(2+0)-s(2)}{0}=\frac{0}{0}=1$
right$\frac{s(2+h)-s(2)}{h}=19.6+4.9h\ \ (h\neq 0),\quad\text{then let }h\to 0$
Drag the step h and watch the secant slope 19.6 + 4.9h; the line stops moving long before h reaches 0.
What a limit is, and what it refuses to look at
Here is what we were left with: the number 19.6 was never one of the values we computed — it was the value they were all heading for. That idea now needs a name, a notation and a definition.
DefinitionDefinition 1.2: The limit of a function
Conditions
$f$ is defined on an open interval around $a$, except possibly at $a$ itself
the same value is approached however $x$ closes in on $a$
$$\boxed{\;\lim_{x\to a}f(x)=L\;}$$
The outputs $f(x)$ can be brought as close to $L$ as anyone demands, by taking $x$ close enough to $a$ — but never equal to $a$.
Three functions near the same point: two of them have the same limit L although one has no value there and the other has the wrong one, and the third has no limit at all.
Looks like this, but is not
This is what a limit is: for $g(x)=\dfrac{x^{2}-9}{x-3}$ the outputs near $3$ close in on $6$, so $\lim_{x\to 3}g(x)=6$.
This looks like the same statement but is not:the limit is just $f(a)$ written slowly. Take $u(x)=x+1$ for every $x\neq 1$ and $u(1)=7$. Then $\lim_{x\to 1}u(x)=2$ while $u(1)=7$. The limit is computed from the neighbours only, and a single stubborn value at the point cannot outvote them.
x
value
side
0.1
0.248457
right
0.01
0.249844
right
0.001
0.249984
right
-0.001
0.250016
left
-0.01
0.250156
left
-0.1
0.251582
left
Both sides are closing in on 0.25, which is strong evidence and not a proof — a table shows finitely many neighbours and a limit is a claim about all of them.
Reading the limit of (√(x+4) − 2)/x at 0 from a table
Estimate $\displaystyle\lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}$ numerically, and say precisely what the estimate is worth.
Given
$f(x)=\dfrac{\sqrt{x+4}-2}{x}$
the table of values above
Find
A candidate value for the limit, and an honest statement of how strong the evidence is.
SolutionCheck that the point itself is closed to us
$f(0)=\frac{\sqrt{4}-2}{0}=\frac{0}{0}$
substitution fails, so $f$ has no value at $0$ and the neighbours are all we have
Read the neighbours from both sides
$f(0.001)=0.249984,\qquad f(-0.001)=0.250016$
the two sides straddle a value instead of drifting apart
$0.249984<0.25<0.250016$
the candidate is trapped between the two sides, which is what a limit looks like numerically
Answer $$\boxed{\lim_{x\to 0}\frac{\sqrt{x+4}-2}{x}\ \text{appears to be}\ \tfrac14}$$
Check
An independent check that does not use the table: if the limit is $\tfrac14$ then $\sqrt{x+4}\approx 2+\tfrac{x}{4}$ for small $x$, and squaring that gives $4+x+\tfrac{x^{2}}{16}$, which is $x+4$ up to a term of size $x^{2}/16$ — far smaller than $x$.
A table can only ever say appears to be. The algebra that turns this into an equality is two blocks away, and it will return exactly 1/4.
Checkpoint
§01.2 — reading a limit and a value off the same graph●○○○○
A graph is drawn on the board. Near $x=2$ the curve rises smoothly towards height 5 from the left and comes down to height 5 from the right, but at $x=2$ itself there is an open circle at height 5 and a filled dot at height 1.
Given
open circle at (2, 5)
filled dot at (2, 1)
the curve approaches height 5 from both sides
Find
(a) State $\lim_{x\to 2}f(x)$ and $f(2)$.
Hint 1/4
Two different questions are being asked about the same picture: what do the neighbours do, and what is drawn at the point.
Hint 2/4
The limit is decided by the approach from both sides; the value is decided by the filled dot.
Hint 3/4
Here both sides approach height 5, and the filled dot sits at height 1.
Hint 4/4
So $\lim_{x\to 2}f(x)=5$ while $f(2)=1$.
Show solutionTake the two questions apart
$\lim_{x\to 2^{-}}f(x)=\lim_{x\to 2^{+}}f(x)=5$
both sides approach the same height, so the limit exists and equals it
$f(2)=1$
the filled dot is the value; the open circle marks a height that is approached but not attained
Consistency check: an open circle and a filled dot in the same vertical line is precisely how a graph draws a function whose limit and value disagree, so the two answers must differ.
⚠ Trusting a table of values
a table that repeats the same number four times feels conclusive, and nothing on the screen warns you otherwise
In every example so far the two sides agreed, and we never had to say so out loud. Now we meet the functions where they do not, and the agreement becomes a condition to be checked.
TheoremTheorem 1.3: Existence test for a two sided limit
Conditions
both one sided limits are computed at the same point $a$
both are finite numbers
$$\boxed{\;\lim_{x\to a}f(x)=L\iff \lim_{x\to a^{-}}f(x)=L\ \text{ and }\ \lim_{x\to a^{+}}f(x)=L\;}$$
A two sided limit exists exactly when both one sided limits exist and are the same number; if either fails, or they differ, there is no two sided limit.
Both branches are well behaved and f(1) is defined, and still there is no limit at 1 — the sides arrive at different heights.
Looks like this, but is not
The limit exists here: $f(x)=\begin{cases}2x,& x<1\\ x+1,& x\ge 1\end{cases}$ has $\lim_{x\to1^{-}}f=2$ and $\lim_{x\to 1^{+}}f=2$, so $\lim_{x\to 1}f(x)=2$.
This looks like the same situation and is not: $g(x)=\begin{cases}x^{2}+1,& x<1\\ 4-x,& x\ge 1\end{cases}$. Both branches are polynomials, $g(1)=3$ is perfectly well defined, and still $\lim_{x\to 1}g(x)$ does not exist — because $2\neq 3$. Niceness of the pieces is not the criterion; agreement at the joint is.
The two sides of a piecewise function at its joint
For $g(x)=\begin{cases}x^{2}+1,& x<1\\ 4-x,& x\ge 1\end{cases}$ find $\lim_{x\to 1^{-}}g(x)$, $\lim_{x\to 1^{+}}g(x)$ and decide whether $\lim_{x\to 1}g(x)$ exists.
Given
$g(x)=x^{2}+1$ for $x<1$
$g(x)=4-x$ for $x\ge 1$
Find
Both one sided limits at 1, and the two sided limit if it exists.
SolutionPick the branch by the side, not by the equals sign
approaching from the right means $x>1$, which is inside the second branch
Apply the existence test
$2\neq 3$
the test asks for equality of the two sides, and it is not met
Answer $$\boxed{\lim_{x\to 1^{-}}g=2,\quad \lim_{x\to1^{+}}g=3,\quad \lim_{x\to 1}g\ \text{does not exist}}$$
Check
Numbers either side confirm the gap rather than a rounding artefact: $g(0.999)=1.998001$ and $g(1.001)=2.999$, a difference of about 1 that does not shrink as we get closer.
Why |x| / x has no limit at 0
Find $\displaystyle\lim_{x\to 0^{+}}\frac{|x|}{x}$ and $\displaystyle\lim_{x\to 0^{-}}\frac{|x|}{x}$, then decide the two sided limit.
Given
$f(x)=\dfrac{|x|}{x}$, defined for $x\neq 0$
Find
Both one sided limits at 0 and the two sided verdict.
SolutionUnpack the absolute value on each side separately
each side is constant, so each one sided limit is that constant
Answer $$\boxed{\lim_{x\to 0}\frac{|x|}{x}\ \text{does not exist}}$$
Check
The function only ever takes the two values $\pm 1$ and takes both in every interval around $0$, so no single number can be approached — this argument never divides by anything.
Choosing the constant c that makes the limit exist
For which value of $c$ does $\displaystyle\lim_{x\to 2}f(x)$ exist, where $f(x)=\begin{cases}x^{2}+c,& x<2\\ 3x-c,& x\ge 2\end{cases}$? What is the limit then?
Given
$f(x)=x^{2}+c$ for $x<2$
$f(x)=3x-c$ for $x\ge 2$
Find
The value of c that makes the two sided limit exist, and the limit.
SolutionWrite both sides in terms of c
$\lim_{x\to 2^{-}}f(x)=2^{2}+c=4+c$
left side, so the first branch; c is a constant and rides along untouched
$\lim_{x\to 2^{+}}f(x)=3\cdot 2-c=6-c$
right side, so the second branch
Impose the existence test and solve
$4+c=6-c$
the test is an equation, not an inequality: this is the only demand existence makes
$2c=2\ \Rightarrow\ c=1$
solving for c rather than for x, because c is the unknown being designed
Report the limit
$\lim_{x\to 2}f(x)=4+c=5$
either side may be used once they are equal, which is a free consistency check
Answer $$\boxed{c=1,\qquad \lim_{x\to 2}f(x)=5}$$
Check
Substituting $c=1$ into the other branch: $6-c=6-1=5$, the same number the left branch gave. Had the two disagreed, the value of $c$ would have been wrong.
Notice what the exam version of this question is really testing: not algebra, but whether you know that existence is an equation between two sides.
Checkpoint
§01.3 — one sided limits with an absolute value in the denominator●●○○○
Absolute values are the cheapest way for an exam to hide a one sided problem inside something that looks two sided. Thirty seconds on this one, and unpack the absolute value before doing anything else.
Given
$h(x)=\dfrac{x^{2}-4}{|x-2|}$
Find
(a) Find $\lim_{x\to 2^{+}}h(x)$ and $\lim_{x\to 2^{-}}h(x)$, then say whether $\lim_{x\to 2}h(x)$ exists.
Hint 1/4
The two sides of 2 make the inside of the absolute value have different signs, so they are two different problems.
Hint 2/4
$|x-2|=x-2$ when $x>2$ and $|x-2|=-(x-2)$ when $x<2$; also $x^{2}-4=(x-2)(x+2)$.
Hint 3/4
On the right: $\dfrac{(x-2)(x+2)}{x-2}=x+2$. On the left: $\dfrac{(x-2)(x+2)}{-(x-2)}=-(x+2)$.
Hint 4/4
So the right limit is $4$, the left limit is $-4$, and the two sided limit does not exist.
Show solutionFactor once, then unpack the absolute value per side
$x^{2}-4=(x-2)(x+2)$
the factor (x − 2) is what the denominator will cancel against
$x>2:\ h(x)=\frac{(x-2)(x+2)}{x-2}=x+2\ \to\ 4$
on the right the inside of the absolute value is positive
The last block had a limit fail because the two sides disagreed about which number to approach. Now we meet the other failure: outputs that do not approach any number at all.
DefinitionDefinition 1.4: Infinite limit and vertical asymptote
Conditions
$f$ is defined on both sides of $a$ (or on one side, for a one sided statement), except possibly at $a$
the outputs pass every bound, rather than settling near one
$$\boxed{\;\lim_{x\to a}f(x)=\infty\ \text{ means the values of }f\text{ exceed every }M>0\text{ for }x\text{ close enough to }a,\ x\neq a\;}$$
Name any height you like, however large, and the graph is above it for every $x$ near enough to $a$. If this happens on either side, the line $x=a$ is a vertical asymptote of the graph.
Two vertical asymptotes with different verdicts, on the same scale: $1/x$ has no limit at 0 because the sides disagree, while $1/x^{2}$ has the informative failure $+\infty$ on both sides.
Looks like this, but is not
This is a vertical asymptote: in $\dfrac{x^{2}-1}{x^{2}-x-2}$ the denominator vanishes at $x=2$ while the numerator equals $3$ there, so the quotient blows up.
This looks like one and is not: the same denominator also vanishes at $x=-1$. But so does the numerator, and $\dfrac{x^{2}-1}{x^{2}-x-2}=\dfrac{(x-1)(x+1)}{(x-2)(x+1)}=\dfrac{x-1}{x-2}$ for $x\neq -1$, whose limit at $-1$ is $\dfrac{-2}{-3}=\dfrac{2}{3}$. A zero denominator is a reason to look, never a conclusion: check the numerator first.
The two sides of (x + 1)/(x − 5) at x = 5
Find $\displaystyle\lim_{x\to 5^{+}}\frac{x+1}{x-5}$ and $\displaystyle\lim_{x\to 5^{-}}\frac{x+1}{x-5}$, and state what happens two sidedly.
Given
$f(x)=\dfrac{x+1}{x-5}$
Find
Both one sided limits at 5 and the two sided verdict.
SolutionCheck the numerator before anything else
$\text{at }x=5:\quad x+1=6\neq 0,\quad x-5=0$
a nonzero over a vanishing quantity cannot be repaired by algebra, so this is a blow up and not a $0/0$
The limit laws, and what 0/0 is really telling you
Tables gave us numbers we believed and pictures gave us numbers we could see. Neither is an argument, and an exam asks for an argument — so we need rules that turn limits into ordinary arithmetic.
RuleRule 1.5: The limit laws
Conditions
$\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=M$ both exist and are finite
for the quotient law, $M\neq 0$
for the root law, $\sqrt[n]{L}$ has to be a real number
If both pieces settle down, then a sum settles on the sum, a product on the product, a power on the power, and a quotient on the quotient — provided the bottom does not settle on zero. The last two lines are the starting stock: a constant approaches itself, and $x$ approaches $a$.
The decision the first line of every limit problem makes: substitute, look at what came out, and let the form choose the repair. Only the zero over zero branch sends you back to the top.
Looks like this, but is not
The laws work forwards: if $\lim_{x\to 0}f=3$ and $\lim_{x\to 0}g=-2$ then $\lim_{x\to 0}(f+g)=1$, with no further checking.
They do not work backwards: take $f(x)=\dfrac1x$ and $g(x)=-\dfrac1x$. Then $f(x)+g(x)=0$ for every $x\neq 0$, so $\lim_{x\to 0}(f+g)=0$ exists — while neither $\lim_{x\to 0}f$ nor $\lim_{x\to 0}g$ exists. The hypothesis of the law is that the pieces have limits; a limit of the combination is no evidence for that.
Limit of 3x² − 5x + 4 at x = 2, with every law named
Compute $\displaystyle\lim_{x\to 2}\left(3x^{2}-5x+4\right)$ and name the law used at each step.
Given
$p(x)=3x^{2}-5x+4$
$a=2$
Find
The limit, with a justification for each line.
SolutionBreak the polynomial into pieces the laws can handle
we rewrite $3-x$ as $-(x-3)$ on purpose, so that the factor matches the one in the denominator
$=\frac{-1}{3x}\qquad (x\neq 3)$
now the shared factor cancels
Substitute
$\lim_{x\to 3}\frac{-1}{3x}=\frac{-1}{9}$
denominator tends to $9\neq 0$, so the quotient law applies
Answer $$\boxed{-\tfrac19}$$
Check
Neighbour check in the original expression: at $x=3.01$, $\frac{1/3.01-1/3}{0.01}=\frac{-0.001107}{0.01}=-0.1107$, close to $-1/9=-0.1111$.
Checkpoint
§01.5 — one factor and cancel, thirty seconds●○○○○
A short one to confirm the reflex: diagnose the form first, and only then decide what to do. If the diagnosis is skipped, the rest of the work is a guess.
Given
$\displaystyle\lim_{x\to -2}\frac{x^{2}-4}{x+2}$
Find
(a) Compute the limit, and name the form you diagnosed on the first line.
Hint 1/4
Before computing anything, decide which of the four cases this is; the case chooses the method.
Hint 2/4
Substituting gives $0/0$, and for polynomials that guarantees a shared factor $(x-a)$ — here $x^{2}-4=(x-2)(x+2)$.
Hint 3/4
Cancel: $\dfrac{(x-2)(x+2)}{x+2}=x-2$ for $x\neq -2$.
Hint 4/4
Then substitute into $x-2$ at $x=-2$: the limit is $-4$.
Show solutionDiagnose, factor, cancel, substitute
$x=-2:\ \frac{0}{0}$
indeterminate, so rewrite rather than conclude
$\frac{(x-2)(x+2)}{x+2}=x-2\quad (x\neq -2)$
the shared factor is the one that vanishes at −2
$\lim_{x\to -2}(x-2)=-4$
a polynomial, so substitution is legal
Answer $$\boxed{-4}$$
Check
At $x=-1.99$ the original expression gives $\frac{3.9601-4}{0.01}=-3.99$, and at $x=-2.01$ it gives $-4.01$.
⚠ Writing 0/0 as the answer
substitution produced something that looks like a result, and no rule was ever stated forbidding it
right$\frac{0}{0}\ \text{is a diagnosis}\Rightarrow \lim_{x\to 3}\frac{(x-3)(x+3)}{(x-3)(x+1)}=\frac{3}{2}$
⚠ Cancelling a term instead of a factor
cancelling looks like a visual operation on symbols rather than an algebraic one on factors
wrong$\frac{x+3}{x}=3$
right$\frac{x+3}{x}=1+\frac{3}{x}\quad\text{(nothing cancels: } x+3 \text{ is a sum, not a product)}$
The Squeeze Theorem: limits with nothing to factor
Every repair so far assumed there was something to cancel. The next function has no common factor, no root to rationalize and no fraction to combine — and it still has a limit.
TheoremTheorem 1.6: The Squeeze Theorem
Conditions
$g(x)\le f(x)\le h(x)$ for all $x$ near $a$, except possibly at $a$ itself
$\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$ — the same $L$ for both
If a function is trapped between two functions that arrive at the same place, it has nowhere else to go, so it arrives there too — and this is proved without ever evaluating the trapped function.
The oscillation of x² sin(1/x) never slows down near 0, but the room it has to oscillate in shrinks to nothing — that is what the two parabolas prove.
Looks like this, but is not
This is a squeeze: $-x^{2}\le x^{2}\sin\frac1x\le x^{2}$, and both bounds tend to $0$, so the middle one does too.
This is not: $-1\le \sin\frac1x\le 1$ is a perfectly true pair of bounds, and it proves nothing about $\lim_{x\to 0}\sin\frac1x$. The theorem needs the two bounds to arrive at the same number, and $-1$ and $1$ are different numbers. That limit really does not exist.
Squeezing x² sin(1/x) at x = 0
Show that $\displaystyle\lim_{x\to 0}x^{2}\sin\frac1x=0$.
Given
$f(x)=x^{2}\sin\dfrac1x$, defined for $x\neq 0$
$-1\le\sin t\le 1$ for every real $t$
Find
The limit at 0, with the bounds written out.
SolutionShow why the product law is unavailable
$\lim_{x\to 0}\sin\frac1x\ \text{does not exist}$
as $x\to0$ the angle $1/x$ runs through every value repeatedly, so the sine keeps taking both $1$ and $-1$
$\Rightarrow \lim x^{2}\cdot \lim \sin\tfrac1x\ \text{is not available}$
the product law has a hypothesis, and one factor fails it
Build the two bounds
$-1\le \sin\frac1x\le 1\qquad (x\neq 0)$
true for every input whatsoever, which is exactly why it survives near a bad point
$-x^{2}\le x^{2}\sin\frac1x\le x^{2}$
multiplying an inequality by $x^{2}\ge 0$ keeps its direction — this step would be wrong with a factor that can be negative
Squeeze
$\lim_{x\to 0}(-x^{2})=0=\lim_{x\to 0}x^{2}$
both bounds are polynomials, so substitution gives their limits
$\Rightarrow \lim_{x\to 0}x^{2}\sin\frac1x=0$
same L at both ends, so the theorem applies
Answer $$\boxed{0}$$
Check
An independent bound rather than a repeat of the algebra: $\left|x^{2}\sin\frac1x\right|\le x^{2}$, so at $|x|<0.01$ every value is within $0.0001$ of $0$ — no oscillation can escape that.
Not one value of the function was computed anywhere in this proof.
A limit forced by two inequalities alone
Suppose $4x-9\le f(x)\le x^{2}-4x+7$ for all $x\ge 0$. Find $\displaystyle\lim_{x\to 4}f(x)$.
Given
$4x-9\le f(x)\le x^{2}-4x+7$ for $x\ge 0$
no formula for f is given at all
Find
The limit of f at 4.
SolutionTake the limits of the two known bounds
$\lim_{x\to 4}(4x-9)=16-9=7$
a polynomial, so substitution
$\lim_{x\to 4}\left(x^{2}-4x+7\right)=16-16+7=7$
also a polynomial, and it lands on the same number
Squeeze
$\Rightarrow\ \lim_{x\to 4}f(x)=7$
the hypotheses are met near 4, so the trapped function has no freedom left
Answer $$\boxed{7}$$
Check
Check that the trap is genuinely tight only at 4: at $x=3$ the bounds are $3$ and $4$, a gap of 1, while at $x=4$ they meet at 7. The pinch is what forces the answer.
This is the shape most exam squeeze questions take: no formula for f, only a sandwich, and the whole answer is the shared value of the two bounds.
Checkpoint
§01.6 — one squeeze in thirty seconds●●○○○
Same pattern as the worked example, one power lower. Careful with the bound you multiply by, since this time it can be negative.
Given
$f(x)=x\cos\dfrac1x$, defined for $x\neq 0$
Find
(a) Find $\lim_{x\to 0}x\cos\dfrac1x$, and write the two bounds you squeezed between.
Hint 1/4
The cosine factor has no limit at $0$, so the product law is out; the question is what traps the product.
Hint 2/4
$-1\le\cos t\le 1$; multiplying an inequality by something that may be negative flips it, so bound the absolute value instead.
Hint 3/4
$\left|x\cos\frac1x\right|\le |x|$, that is $-|x|\le x\cos\frac1x\le |x|$, and $\lim_{x\to0}|x|=0$.
Every computation in the last two blocks ended the same way — rewrite until substitution becomes legal. Continuity is the name of the property that makes it legal, and it is what the whole chapter has been circling.
DefinitionDefinition 1.7: Continuity at a point
Conditions
$f(a)$ is defined — that is, $a$ is in the domain of $f$
The value the function heads for and the value it actually takes are the same number — one equation that quietly demands all three conditions above.
The three ways continuity fails at a point, on one graph: a missing value at 1, disagreeing sides at 3, and no limit at all at 5. Only the first can be repaired.
Looks like this, but is not
This function is continuous at 3: $g(x)=x+3$, since $\lim_{x\to 3}g(x)=6=g(3)$ — all three conditions hold.
This one looks identical and is not continuous at 3: $f(x)=\dfrac{x^{2}-9}{x-3}$. It equals $x+3$ at every point where it is defined, and its limit at $3$ is $6$ — but $f(3)$ does not exist, so the first condition already fails. Cancelling changed the domain, and the domain is part of the function. The defect is removable: defining $f(3)=6$ produces a continuous function, but it is a new function.
Testing continuity of a three part piecewise function at x = 2
Is $f(x)=\begin{cases}x^{2}+1,& x<2\\ 5,& x=2\\ 3x-1,& x>2\end{cases}$ continuous at $x=2$? Check all three conditions.
Given
$f(x)=x^{2}+1$ for $x<2$
$f(2)=5$
$f(x)=3x-1$ for $x>2$
Find
Whether f is continuous at 2, condition by condition.
SolutionCondition 1: is the value there?
$f(2)=5$
given explicitly by the middle line, so the point is in the domain
Condition 2: does the limit exist?
$\lim_{x\to 2^{-}}f(x)=2^{2}+1=5$
left side, first branch, substitution legal because it is a polynomial
$\lim_{x\to 2^{+}}f(x)=3(2)-1=5$
right side, third branch
$\Rightarrow \lim_{x\to 2}f(x)=5$
the sides agree, so the two sided limit exists
Condition 3: do they agree?
$\lim_{x\to 2}f(x)=5=f(2)$
the equation in the definition is satisfied
Answer $$\boxed{\text{Yes: } f \text{ is continuous at } 2}$$
Check
Change one number to see that the test has teeth: with $f(2)=4$ instead, conditions 1 and 2 still hold and condition 3 fails, giving a .
Choosing k so that the function is continuous at x = 4
Find $k$ so that $f(x)=\begin{cases}\dfrac{x^{2}-16}{x-4},& x\neq 4\\ k,& x=4\end{cases}$ is continuous at $4$.
Given
$f(x)=\dfrac{x^{2}-16}{x-4}$ for $x\neq 4$
$f(4)=k$
Find
The value of k that makes f continuous at 4.
SolutionCompute the limit, which does not involve k
the limit only sees $x\neq 4$, which is exactly the branch that carries no $k$
$\lim_{x\to 4}f(x)=4+4=8$
substitution in the repaired expression
Impose the definition
$\text{continuity at }4\iff \lim_{x\to 4}f(x)=f(4)=k$
the definition read as an equation for the unknown value
$\Rightarrow k=8$
there is exactly one such $k$, because a limit is a single number
Answer $$\boxed{k=8}$$
Check
With $k=8$ the whole function is just $x+4$ rewritten, and $x+4$ is a polynomial — continuous everywhere, as required.
Removable discontinuities are the only ones that can be repaired this way, and this computation is the repair: the limit hands you the value the point should have had.
Checkpoint
§01.7 — which of the three conditions fails●●○○○
Thirty seconds, and the question is diagnostic rather than computational: continuity can fail in three distinct places, and naming which one fails is what decides whether the defect can be repaired.
Given
$f(x)=\dfrac{x-2}{x^{2}-4}$
the point $a=2$
Find
(a) Which condition of the definition fails at 2, and can the discontinuity be repaired?
Hint 1/4
Go through the three conditions in order rather than looking at the graph; the first one that fails is the diagnosis.
Hint 2/4
Condition 1 is about the domain; conditions 2 and 3 are about the limit. Also $x^{2}-4=(x-2)(x+2)$.
Hint 3/4
At $x=2$: substitution gives $0/0$, so $f(2)$ is undefined. Cancelling, $\frac{x-2}{(x-2)(x+2)}=\frac{1}{x+2}\to\frac14$.
Hint 4/4
So condition 1 fails while the limit exists: the discontinuity at 2 is removable, repaired by defining $f(2)=\tfrac14$.
Show solutionWalk the three conditions
$f(2)=\frac{0}{0}\ \text{undefined}$
condition 1 fails, so f is certainly not continuous at 2
a single missing value is the only defect, which is what removable means
Answer $$\boxed{\text{condition 1 fails; removable with } f(2)=\tfrac14}$$
Check
Neighbour check: $f(2.01)=\frac{0.01}{0.0401}=0.2494$ and $f(1.99)=\frac{-0.01}{-0.0399}=0.2506$, both near $0.25$.
⚠ Turning the break at one point into a verdict on the whole function
the blow up at 0 is dramatic, so a local report gets promoted into a global one
wrong$\tfrac1x\ \text{has an infinite discontinuity at }0\ \Rightarrow\ \tfrac1x\ \text{is not continuous}$
right$\tfrac1x\ \text{has an infinite discontinuity at }0,\ \text{and is continuous at every point of its domain}$
⚠ Cancelling and then claiming continuity
after cancelling, the formula on the page really is continuous — but it is a different function
wrong$\frac{x^{2}-9}{x-3}=x+3\ \Rightarrow\ \text{continuous at }3$
right$\frac{x^{2}-9}{x-3}=x+3\ \text{only for } x\neq 3;\ \text{at }3\ \text{the value is missing, so it is discontinuous (removably)}$
Computing a limit: the order of attack
Every limit problem in this section, and the first line of every limit problem for the rest of the course.
Substitute
Put $x=a$ into the formula. This is a diagnostic move, not yet an answer.
A number came out
Then that number is the limit, provided the formula is one of the standard types at a point of its domain — see the substitution licence below. Stop here.
$0/0$ came out
Rewrite until the vanishing factor is visible and cancels: factor polynomials, multiply by the conjugate when a root is in the way, combine stacked fractions, unpack any absolute value. Then go back to step 1.
Nonzero over $0$ came out
No rewrite can help. Take the sides one at a time, get the sign of the denominator on each, and report $+\infty$ or $-\infty$ per side.
The rule changes at $a$
If the formula is piecewise, or contains an absolute value, or only exists on one side, do the two sides separately and compare them with the existence test.
Something bounded is oscillating
If a factor stays between fixed bounds while the rest tends to $0$, do not use the product law — build a sandwich and squeeze.
Where it goes wrong
Stopping at step 3 and writing $0/0$ as the answer.
Doing a two sided computation on a problem whose formula changes at the point — the answer will look clean and be wrong.
Cancelling a factor without recording that the cancellation is valid only away from the point.
Getting the sign of an infinite limit right
Whenever substitution gives a nonzero number over $0$, and the question asks for one side or for the shape of the graph.
Confirm the numerator
Check that the numerator does not vanish at $a$. If it does, this is a $0/0$ problem and there may be no asymptote at all.
Factor the denominator
Write the denominator as a product; only the factor that vanishes at $a$ matters for the sign, the rest can be evaluated.
Sign each factor on the chosen side
For $x\to a^{+}$ the factor $(x-a)$ is a small positive number; for $x\to a^{-}$ it is a small negative number; $(x-a)^{2}$ is positive on both sides.
Combine and report
Count the negative signs among the numerator and the surviving factors: an even count gives $+\infty$, an odd count gives $-\infty$. Report each side separately, then compare.
Where it goes wrong
Forgetting that an even power of the vanishing factor keeps the same sign on both sides, so the two sided infinite statement is then legitimate.
Ignoring the sign of the numerator when it is negative.
When substitution is legal: the direct substitution property
At step 2 of the recipe, to justify writing the answer immediately instead of doing any work.
Identify the build
Is $f$ assembled from polynomials, rational functions, roots, and trigonometric functions by adding, multiplying, dividing and composing?
Check membership
Is $a$ a point of the domain of that assembled formula — in particular, no zero denominator and no negative quantity under an even root?
Use the property
If both hold, $f$ is continuous at $a$, so $\displaystyle\lim_{x\to a}f(x)=f(a)$ and the answer is the substitution you already did.
If not
If either fails, you are in one of the repair cases of the recipe. The failure of substitution is information about which repair to use, never a verdict on the limit.
Where it goes wrong
Using the licence at a point outside the domain — the property says nothing there.
Applying it to a piecewise formula at the joint, where the function is not one of the standard builds.
the numerator vanishes too, so the trouble may be removable
$\frac{(x-2)(x+2)}{x-2}=x+2\ (x\neq 2)$
the shared factor cancels
$\lim_{x\to 2}(x+2)=4$
substitution in the repaired expression is legal
Answer $$\boxed{4}$$
Check
At $x=2.001$ the original gives $4.001$ — finite and near 4, not large.
Nonzero over zero: (x² + 4)/(x − 2) at x = 2
Compute $\displaystyle\lim_{x\to 2^{+}}\frac{x^{2}+4}{x-2}$ and $\displaystyle\lim_{x\to 2^{-}}\frac{x^{2}+4}{x-2}$.
Given
$\dfrac{x^{2}+4}{x-2}$
Find
Both one sided limits at 2.
SolutionDiagnose, then sign each side
$x=2:\ \frac{8}{0}$
the numerator does not vanish, so no cancellation is possible and nothing can be repaired
$x\to 2^{+}:\ \frac{8}{0^{+}}\to+\infty$
just right of 2 the denominator is a small positive number
$x\to 2^{-}:\ \frac{8}{0^{-}}\to-\infty$
just left of 2 it is a small negative number
Answer $$\boxed{+\infty\ \text{and}\ -\infty;\ \text{no two sided limit}}$$
Check
At $x=2.001$ the value is $8004$ and at $x=1.999$ it is $-7996$ — enormous and of opposite signs.
The two expressions differ by a single sign inside the numerator, and that sign decides whether the answer is a small whole number or a pair of infinities.
How to tell them apart
Evaluate the numerator at $a$ first: zero on top means rewrite and expect a finite answer; nonzero on top means take the sides and report signed infinities.
the sign flip is still there, but both branches now head for the same number
$0=0$
the existence test is satisfied
Answer $$\boxed{0}$$
Check
Equivalently $\frac{x^{2}}{|x|}=|x|$ for $x\neq0$, and $|x|\to 0$ — a second route to the same answer.
Both functions flip sign at 0 and both are undefined there, yet one has no limit and the other has a perfectly ordinary one.
How to tell them apart
Unpack the absolute value on each side and then compare the two branches: if the flip leaves two different numbers, the limit dies; if both branches head to the same number, the flip is irrelevant.
Scaffolding comes off
The common skeleton
Substitute the point and name the form you get.
Let the form choose the repair: factor, conjugate, combine, or unpack an absolute value.
Do the algebra, and write down that it is valid only for $x\neq a$.
Substitute into the repaired expression.
Check the answer against a neighbour of the point.
1 · fully worked
Rung 1 — worked in full: (x² + 2x − 3)/(x² − 1) at x = 1
cancelling is legal off the point, and off the point is where the limit lives
Step 4 — substitute into the repaired expression
$\lim_{x\to 1}\frac{x+3}{x+1}=\frac{4}{2}=2$
the denominator tends to $2\neq 0$, so the quotient law applies
Answer $$\boxed{2}$$
Check
Step 5, with a neighbour: at $x=1.01$ the original expression gives $\frac{0.0401}{0.0201}=1.995$.
2 · you write the reasoning
Now an easier one, but you supply the reasons. Compute $\displaystyle\lim_{x\to 4}\frac{x^{2}-16}{x-4}$ and write, in the empty column, why each line is allowed.
reasoning
Substitution is a diagnosis: the form is indeterminate, so nothing has been decided yet and a rewrite is required.
reasoning
The form $0/0$ guarantees that both polynomials vanish at 4, so both contain the factor $(x-4)$ — this is the factor to expose.
reasoning
Cancelling is valid only for $x\neq 4$, and a limit at 4 never looks at $x=4$, so no information is lost.
reasoning
The repaired expression is a polynomial, so the substitution licence applies and the limit is its value at 4.
3 · find the buried error
Harder, and this time the work is done for you — badly. Two of the four steps below are wrong. Find both. The problem: compute $\displaystyle\lim_{x\to 2^{-}}\frac{x^{2}-3x+2}{|x-2|}$.
the two buried errors (2)
⚠ step 2
The branch of the absolute value was chosen by closeness to 2 instead of by side. Approaching from the left means $x<2$, so $x-2<0$ and $|x-2|=-(x-2)$.
The bars look like they can be removed once x is near the point, and the minus sign has no visible cause on the page.
right
For $x<2$: $\dfrac{(x-1)(x-2)}{-(x-2)}=-(x-1)$, so the left hand limit is $-(2-1)=-1$.
⚠ step 4
A one sided computation was reported as a two sided limit, and continuity was claimed at a point where the function is not even defined.
Once a number appears, the habit is to write the plain limit symbol; and continuity gets asserted from the existence of a limit alone, skipping the domain condition.
right
Only the left hand limit was computed, so only $\lim_{x\to 2^{-}}$ may be reported. The right hand limit is $+ (2-1)=1$, so the two sided limit does not exist; and $f(2)$ is undefined, so continuity at 2 is impossible.
4 · the bare problem
§01.3 — one sided limit with an absolute value, no scaffolding●●●○○
Same skeleton, no help this time. Diagnose the form, choose the repair, keep track of the side, and check your answer against a neighbour before looking.
Two things need deciding before any algebra: which form this is, and which side you are on.
Hint 2/4
$x^{2}-9=(x-3)(x+3)$, and for $x>3$ the inside of the absolute value is positive, so $|x-3|=x-3$.
Hint 3/4
Then $\dfrac{(x-3)(x+3)}{x-3}=x+3$ for $x>3$, and the point is $a=3$.
Hint 4/4
Substituting into $x+3$ gives $6$.
Show solutionDiagnose, unpack, cancel, substitute
$x=3:\ \frac{0}{0}$
indeterminate, so a rewrite is required
$x>3\ \Rightarrow\ |x-3|=x-3$
the side fixes the branch; from the left the answer would carry the opposite sign
$\frac{(x-3)(x+3)}{x-3}=x+3\qquad (x\neq 3)$
cancelling the shared factor
$\lim_{x\to 3^{+}}(x+3)=6$
polynomial, so substitution is licensed
Answer $$\boxed{6}$$
Check
Neighbour on the correct side: at $x=3.01$ the original gives $\frac{0.0601}{0.01}=6.01$. On the wrong side, $x=2.99$ would give $-5.99$, which is how you catch a side error.
Full exam-style question
Exam format: a function with a jump, an asymptote and a hidden holeexam format
A full length limits question sets one piecewise function and then walks four points of interest past it, with the parts getting harder as they go. The marks are in the reasoning, and every part is decided by one of four diagnoses: a number, zero over zero, nonzero over zero, or a rule that changes at the point.
(a) the two one sided limits at 3; (b) whether the limit at 3 exists and whether f is continuous there; (c) the two one sided limits at 4 and the asymptote; (d) all discontinuities in [0, 6], classified.
the sign of the denominator is what the side decides
$\Rightarrow x=4\ \text{is a vertical asymptote}$
one side is enough for the asymptote, and here both blow up
(d) All discontinuities in [0, 6]
$x=3:\ \text{jump discontinuity}$
both one sided limits exist and are finite but differ, which is the definition of a jump; not repairable by any single value
$x=4:\ \text{infinite discontinuity}$
the limit fails by blowing up, so again no value can repair it
$\text{elsewhere on }[0,6]:\ \text{continuous}$
each branch is a rational function and no other point of the interval kills a denominator
Answer $$\boxed{\text{(a) }5\ \text{and}\ -6;\quad\text{(b) no limit, not continuous};\quad\text{(c) }-\infty,\ +\infty,\ x=4;\quad\text{(d) jump at }3,\ \text{infinite at }4}$$
Check
Spot check every claim with one neighbour each: $f(2.99)=4.99$, $f(3.01)=\frac{6.02}{-0.99}=-6.08$, $f(3.99)=\frac{7.98}{-0.01}=-798$, $f(4.01)=\frac{8.02}{0.01}=802$.
Four parts, four diagnoses, and only one of them needed algebra.
What each skipped step costs. Skip the factoring in (a) and you report $0/0$ or, worse, a wrong finite number — and every later part that quotes the left limit inherits it. Skip the sides in (c) and you write a single $\infty$, which loses the entire sign discussion the part exists to test. Skip the domain check in (d) and you call the jump removable, which is the single most common error on this question type: a jump is never repairable, because no choice of one value can equal two different one sided limits.
Practice
A · concept 4 questions
1§01.2 — value at the point versus limit at the point●○○○○
A classmate is revising and says: a function has to be defined at a point before it can have a limit there, otherwise there is nothing to take the limit of. Decide whether the statement below, which is his claim written precisely, is true or false.
Given
Claim: if $f(a)$ is undefined, then $\lim_{x\to a}f(x)$ cannot exist.
Find
(a) True or false, with a reason or a counterexample.
Hint 1/4
Ask which points the definition of a limit actually looks at.
Hint 2/4
The definition talks about $x$ close to $a$ with $x\neq a$; the point itself is deliberately excluded.
Hint 3/4
Test it on $f(x)=\dfrac{x^{2}-4}{x-2}$: $f(2)$ is undefined and the values near 2 close in on 4.
Hint 4/4
So the claim is false, and the counterexample is one line long.
The instantaneous speed computation was the same shape: the difference quotient is undefined at $h=0$ and still has the limit 19.6.
2§01.3 — what the existence test actually demands●●○○○
An answer sheet reads: both one sided limits exist here, therefore the two sided limit exists and equals one of them. The reasoning is stated below as a general claim.
Given
Claim: if $\lim_{x\to a^{-}}f(x)$ and $\lim_{x\to a^{+}}f(x)$ both exist, then $\lim_{x\to a}f(x)$ exists.
Find
(a) True or false, with a reason or a counterexample.
Hint 1/4
Write the existence test out in full and count how many demands it makes.
Hint 2/4
$\lim_{x\to a}f=L$ requires both one sided limits to exist and to be the same number $L$.
Hint 3/4
Test it on $f(x)=\dfrac{|x|}{x}$ at $0$: the right limit is 1, the left limit is $-1$, both existing.
Hint 4/4
So the claim is false: existence of both sides is necessary but not sufficient.
Show solutionSplit the test into its two demands
$\text{existence of both sides}\ \wedge\ \text{equality}$
both exist, and the two sided limit still does not
Answer $$\boxed{\text{False}}$$
Check
The piecewise example from the figure does the same job with different numbers: 2 and 3 both exist, and the limit does not.
3§01.4 — what an infinite limit statement claims●●○○○
A grader writes 'the limit exists and equals $+\infty$' on one script and 'the limit does not exist, it is $+\infty$' on another, and marks both correct. Only one of the statements below explains why.
Given
The statement $\lim_{x\to a}f(x)=+\infty$
Find
(a) Which reading is correct?
Hint 1/4
Ask what kind of object $+\infty$ is before asking what the equation claims.
Hint 2/4
$\infty$ is not a real number, so an equation ending in $\infty$ cannot be reporting a value in the usual sense.
Hint 3/4
In $\lim_{x\to 0}\frac{1}{x^{2}}=+\infty$ the outputs at $x=0.01$ and $x=0.001$ are $10^{4}$ and $10^{6}$ — passing every bound.
Hint 4/4
So the statement describes the manner of failure, and no finite limit exists.
Show solutionClassify the symbol, then the claim
$+\infty\notin\mathbb{R}$
so the equation is not reporting a real value
$\forall M>0\ \exists\ \text{neighbourhood of } a:\ f(x)>M$
the actual content of the notation
Answer $$\boxed{\text{a description of unbounded growth, not a value}}$$
Check
Consistency test: if $+\infty$ were a value, the difference law would let us write $\infty-\infty=0$, which fails for $\frac{1}{x^{2}}-\frac{1}{x^{4}}$ at 0.
4§01.5 — the limit laws only run forwards●●●○○
On a homework a student argues: the sum has a limit, so each part must have one too, otherwise where would the sum's limit come from. The claim is written out below.
Given
Claim: if $\lim_{x\to a}\bigl[f(x)+g(x)\bigr]$ exists, then $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ exist.
Find
(a) True or false, with a reason or a counterexample.
Hint 1/4
Read the sum law and identify which part is the hypothesis and which is the conclusion.
Hint 2/4
The law says: if $\lim f$ and $\lim g$ exist, then $\lim(f+g)$ is their sum. It claims nothing in the other direction.
Hint 3/4
Take $f(x)=\frac1x$ and $g(x)=-\frac1x$ at $a=0$: then $f(x)+g(x)=0$ for every $x\neq 0$.
Hint 4/4
So $\lim_{x\to0}(f+g)=0$ exists while neither piece has a limit, and the claim is false.
A second, different counterexample confirms it is not a fluke of that pair: $f(x)=\frac{|x|}{x}$ and $g(x)=-\frac{|x|}{x}$ at 0.
B · computation 8 questions
1§01.1 — a difference quotient in disguise●●○○○
This is the limit that will define the derivative in a few weeks, met here as an ordinary algebra problem. It looks harder than it is because the variable is called h.
Given
$\displaystyle\lim_{h\to 0}\frac{(3+h)^{2}-9}{h}$
Find
(a) Compute the limit.
Hint 1/4
Diagnose the form first; the letter used for the variable changes nothing.
Hint 2/4
Substituting $h=0$ gives $0/0$, so expand the square and look for the factor $h$ to cancel.
Hint 3/4
$(3+h)^{2}-9=9+6h+h^{2}-9=6h+h^{2}=h(6+h)$, and the denominator is $h$.
Hint 4/4
So the quotient is $6+h$ for $h\neq0$, whose limit is 6.
Show solutionExpand, factor, cancel, substitute
$h=0:\ \frac{0}{0}$
indeterminate, so rewrite
$(3+h)^{2}-9=6h+h^{2}=h(6+h)$
expanding is what exposes the factor h; the constant 9 cancels by design
$\frac{h(6+h)}{h}=6+h\qquad(h\neq0)$
cancellation valid away from the point
$\lim_{h\to0}(6+h)=6$
polynomial in h, so substitution is licensed
Answer $$\boxed{6}$$
Check
At $h=0.001$ the original expression gives $\frac{9.006001-9}{0.001}=6.001$.
2§01.5 — factor and cancel with a sign to watch●●○○○
A routine factor and cancel, with one trap: the point is negative, so a careless sign in the last substitution changes the answer entirely.
Decide which of the four diagnoses applies before doing anything else.
Hint 2/4
Numerator at $-2$ is $5\neq 0$ and the denominator vanishes, so this is a blow up and the only question is the sign.
Hint 3/4
For $x\to-2^{+}$ we have $x>-2$, so $x+2$ is a small positive number while the numerator stays near 5.
Hint 4/4
A fixed positive number over a small positive number grows without bound, so the limit is $+\infty$.
Show solutionClassify, then sign
$x=-2:\ \frac{5}{0}$
nonzero over zero, so no algebra can repair it
$x\to-2^{+}\Rightarrow x+2\to 0^{+}$
approaching from the right means x is larger than −2
$\frac{5}{0^{+}}\to+\infty$
positive over small positive
Answer $$\boxed{+\infty}$$
Check
At $x=-1.999$ the value is $\frac{4.996}{0.001}=4996$, and it grows as the point is approached. From the other side, $x=-2.001$ gives $-5004$, which is why the side had to be respected.
6§01.7 — a difference of two blow ups that is not a blow up●●●○○
Each of the two terms separately runs off to infinity at 1. Subtracting one infinity from another is meaningless, so the expression has to be combined into a single fraction before anything can be diagnosed.
the numerator now vanishes at 1 too, which is what saves the expression
$\lim_{x\to1}\frac{1}{x+1}=\frac12$
denominator tends to $2\neq0$
Answer $$\boxed{\tfrac12}$$
Check
At $x=1.001$: $\frac{1}{0.001}-\frac{2}{0.002001}=1000-999.5=0.4998$, near $\tfrac12$ — and notice the two huge terms nearly cancelling.
7§01.6 — squeeze with a fourth power●●●○○
An oscillating factor whose argument blows up, multiplied by something that dies. The product law is unavailable, so the answer has to be forced by bounds.
Given
$\displaystyle\lim_{x\to 0}x^{4}\cos\frac{2}{x}$
Find
(a) Compute the limit, showing the sandwich you used.
Hint 1/4
The cosine factor has no limit at $0$, so ask what bounds it instead.
Hint 2/4
$-1\le\cos t\le 1$ for every $t$, including $t=2/x$; and $x^{4}\ge 0$ for every $x$, so multiplying keeps the direction.
Hint 3/4
$-x^{4}\le x^{4}\cos\frac{2}{x}\le x^{4}$, and $\lim_{x\to0}(-x^{4})=\lim_{x\to0}x^{4}=0$.
Hint 4/4
Both walls arrive at 0, so the trapped function does too.
Show solutionBound and squeeze
$-1\le\cos\frac2x\le1\qquad(x\neq0)$
true whatever the argument does, which is why it survives near the bad point
$-x^{4}\le x^{4}\cos\frac2x\le x^{4}$
multiplying by $x^{4}\ge0$ preserves the direction of both inequalities
the numerator becomes $x-5$ exactly, which is the factor the denominator carries
$=\frac{1}{\sqrt{x-1}+2}\qquad(x\neq5)$
cancelling away from the point
$\lim_{x\to5}\frac{1}{\sqrt{x-1}+2}=\frac{1}{4}$
denominator tends to $4\neq0$
Answer $$\boxed{\tfrac14}$$
Check
At $x=5.01$ the original gives $\frac{2.0025-2}{0.01}=0.24984$, and $1/4=0.25$.
C · exam level 4 questions
1§01.5 — the laws applied to limits you are given rather than functions you can see●●●○○
An exam favourite: no formulas at all, only two limits, so the only thing being tested is whether you know each law and its hypothesis. Part (d) is the one that separates scripts.
(d) $\displaystyle\lim_{x\to 2}\frac{f(x)}{g(x)+2}$ — decide whether the given data determines it, and say why.
Hint 1/4
For each part, first ask which law you want to use and whether its hypothesis is satisfied by the given data.
Hint 2/4
Sum, constant multiple, power, product, root laws all need the pieces to have limits, which they do; the quotient law additionally needs the denominator's limit to be nonzero.
Hint 3/4
With $\lim f=4$ and $\lim g=-2$: (a) $3(4)-(-2)^{2}$; (b) $\frac{4\cdot(-2)}{4+(-2)}$; (c) $\sqrt{4+5}$; (d) denominator $\to -2+2=0$.
Hint 4/4
So (a) $=8$, (b) $=-4$, (c) $=3$, and (d) is not determined: the quotient law does not apply and the data says nothing about the sign of $g+2$ near 2.
Show solution(a) constant multiple, power, difference
the quotient law applies because the denominator's limit is $2\neq0$
$=\frac{-8}{2}=-4$
arithmetic
(c) root law
$\lim\sqrt{f+5}=\sqrt{\lim f+5}=\sqrt{9}=3$
legal because $9\ge 0$, so the root is a real number
(d) the law that refuses
$\lim\left(g(x)+2\right)=-2+2=0$
the hypothesis of the quotient law fails, so the law says nothing
$\text{numerator}\to 4\neq 0\Rightarrow\ \text{the quotient is unbounded near }2$
so no finite limit is possible, but the sign and even the existence of a one sided infinity depend on how g approaches −2
Answer $$\boxed{8,\ -4,\ 3,\ \text{(d) not determined}}$$
Check
Part (d) is settled by two examples with the same data: $g(x)=-2+(x-2)$ gives $\lim_{x\to2^{+}}=+\infty$, while $g(x)=-2+(x-2)^{2}\sin\frac{1}{x-2}$ makes even the one sided behaviour oscillate. Same hypotheses, different outcomes.
Part (d) is why the quotient law is always quoted with its condition attached: without it the law is simply false.
2§01.3 — two joints, two unknowns●●●●○
The full version of the matching problem: three branches, two joints, and two constants to be chosen so that both joints work. It is a system of two linear equations dressed up as a limit question.
(a) Write $\lim_{x\to2^{-}}f(x)$ and $\lim_{x\to2^{+}}f(x)$ in terms of $a$ and $b$.
(b) Write $\lim_{x\to5^{-}}f(x)$ and $\lim_{x\to5^{+}}f(x)$ in terms of $a$ and $b$.
(c) Find $a$ and $b$ so that both two sided limits exist.
(d) State the value of each of those two limits.
Hint 1/4
Each joint contributes one equation, and each equation comes from the existence test at that joint.
Hint 2/4
At a joint $c$, existence means $\lim_{x\to c^{-}}f(x)=\lim_{x\to c^{+}}f(x)$; the branch is chosen by the side, never by where the equality case is written.
Hint 3/4
Left at 2: $2^{2}-1=3$. Right at 2: $2a+b$. Left at 5: $5a+b$. Right at 5: $3-5=-2$.
Hint 4/4
Solving $2a+b=3$ and $5a+b=-2$ gives $a=-\tfrac53$, $b=\tfrac{19}{3}$, with limits $3$ at $x=2$ and $-2$ at $x=5$.
Show solution(a) and (b): the four one sided limits
$\lim_{x\to2^{-}}f(x)=2^{2}-1=3$
left of 2 the first branch is in force
$\lim_{x\to2^{+}}f(x)=2a+b$
right of 2 the middle branch is in force, and a, b ride along
$\lim_{x\to5^{-}}f(x)=5a+b$
just left of 5 we are still inside the middle branch
$\lim_{x\to5^{+}}f(x)=3-5=-2$
right of 5 the last branch takes over
(c): two existence tests give two equations
$2a+b=3$
existence at the first joint
$5a+b=-2$
existence at the second joint
$(5a+b)-(2a+b)=3a=-5\ \Rightarrow\ a=-\tfrac53$
subtracting eliminates b, which is the cheaper unknown to remove here
$b=3-2a=3+\tfrac{10}{3}=\tfrac{19}{3}$
back substitution into the first equation
(d): report the limits
$\lim_{x\to2}f(x)=3,\qquad \lim_{x\to5}f(x)=-2$
each equals the branch value that did not contain the unknowns, which is the quickest way to state them
Substitute back into the branch that was not used to state each answer: $2a+b=-\tfrac{10}{3}+\tfrac{19}{3}=3$ and $5a+b=-\tfrac{25}{3}+\tfrac{19}{3}=-2$. Both match.
3§01.4 and 01.7 — one rational function, every diagnosis in the section●●●●○
One function, two suspicious points, and two completely different verdicts. This is the shape of the standard exam question on asymptotes and continuity, and the marks are for telling the two points apart.
Given
$f(x)=\dfrac{x^{2}-x-2}{x^{2}-4}$
Find
(a) Factor and simplify, stating where the simplification is valid.
(b) Find $\lim_{x\to2}f(x)$.
(c) Find $\lim_{x\to-2^{-}}f(x)$ and $\lim_{x\to-2^{+}}f(x)$, and give the vertical asymptote.
(d) List every point where f fails to be continuous, classify each, and say which can be repaired.
Hint 1/4
Two points make the denominator vanish; the numerator decides which of them is a hole and which is an asymptote.
Hint 2/4
Factor: $x^{2}-x-2=(x-2)(x+1)$ and $x^{2}-4=(x-2)(x+2)$. A shared factor means a hole, an unmatched one means a blow up.
Hint 3/4
So $f(x)=\dfrac{x+1}{x+2}$ for $x\neq2$. At $x=2$: $\tfrac34$. At $x=-2$: numerator $\to-1\neq0$, denominator $\to0$.
Hint 4/4
Hence $\lim_{x\to2}f=\tfrac34$ (a hole), $\lim_{x\to-2^{-}}f=+\infty$, $\lim_{x\to-2^{+}}f=-\infty$, and $x=-2$ is the vertical asymptote.
$x=2:\ \text{removable, repair with } f(2)=\tfrac34$
the limit exists and only the value is missing
$x=-2:\ \text{infinite, not repairable}$
no value can be assigned because the limit itself does not exist
Answer $$\boxed{\text{hole at }2\ \left(\tfrac34\right);\quad \text{asymptote } x=-2:\ +\infty\ \text{from the left},\ -\infty\ \text{from the right}}$$
Check
Four neighbours, one per claim: $f(1.99)=0.74937$, $f(2.01)=0.75062$, $f(-2.001)=1001$, $f(-1.999)=-999$.
Notice that the sign pattern at −2 is the opposite of the usual one, because the numerator is negative there. Reading the signs off a memorised picture instead of the actual numerator is the standard way to lose these marks.
4§01.6 — how much a sandwich can force●●●●○
A squeeze question with a sting in the tail. The inequality is stated for all x, not merely for x near 2, and part (b) asks what that stronger hypothesis buys you.
Given
$2x-1\le f(x)\le x^{2}-2x+3$ for every real $x$
Find
(a) Find $\lim_{x\to2}f(x)$.
(b) Can $f(2)$ be determined? If so, find it.
(c) Is $f$ continuous at 2?
(d) Which answers change if the inequality is only assumed for $x\neq 2$?
Hint 1/4
Separate what the inequality says about neighbours of 2 from what it says at 2 itself.
Hint 2/4
The Squeeze Theorem uses the inequality near $a$ and needs both bounds to share a limit; an inequality that also holds at $a$ additionally pins the value there.
Hint 3/4
Bounds at $x=2$: $2(2)-1=3$ and $2^{2}-2(2)+3=3$. Both bounds have limit 3 at 2, and both equal 3 at 2.
Hint 4/4
So $\lim_{x\to2}f(x)=3$; the inequality at $x=2$ forces $3\le f(2)\le3$, so $f(2)=3$; hence f is continuous at 2. Without the point itself, only the limit survives.
with the value free, f could be continuous or have a removable discontinuity
Answer $$\boxed{3,\quad f(2)=3,\quad \text{continuous},\quad \text{(d) only the limit survives}}$$
Check
A concrete f obeying both bounds: $f(x)=x^{2}-2x+3$ itself gives $f(2)=3$; so does $f(x)=2x-1$. Two very different functions, the same forced value at 2.
D · interleaved 4 questions
1mixed practice — decide the method yourself●●●●○
From here on the type of each problem is deliberately not announced. Look at the expression, run the diagnosis, and commit to a method before you start computing; deciding what kind of problem you are holding is the skill being trained.
Given
$\displaystyle\lim_{x\to 0}\frac{|x|}{x^{2}+x}$
Find
(a) Compute the limit, or show that it does not exist.
Hint 1/4
Before any algebra, look for anything in the expression whose behaviour depends on which side of the point you are on.
Hint 2/4
An absolute value must be unpacked per side: $|x|=x$ for $x>0$ and $|x|=-x$ for $x<0$. The denominator factors as $x(x+1)$.
At $x=2.001$ the original gives $\frac{0.012006}{0.004001}=3.0008$.
The type was: plain factor and cancel, disguised by a cubic that needs the less familiar identity.
3mixed practice — decide the method yourself●●●●○
This one is designed to be misdiagnosed. A denominator that vanishes and a one sided arrow both suggest an infinite limit; check the numerator before you commit to that story.
the numerator carries the factor twice, so one copy survives the cancellation
$\lim_{x\to1^{-}}\frac{x-1}{x^{2}}=\frac{0}{1}=0$
the side makes no difference once the expression has been repaired
Answer $$\boxed{0}$$
Check
At $x=0.99$ the original gives $\frac{0.0001}{-0.009801}=-0.0102$, small and heading to 0 — not large, so the asymptote reading would have been wrong.
The type was: factor and cancel, with a repeated factor. The one sided arrow was a decoy, and a vanishing denominator on its own never decides anything.
4mixed practice — decide the method yourself●●●●○
A tank is draining. Its volume after t minutes is modelled by $V(t)=200-8t+0.08t^{2}$ litres, valid for $0\le t\le 50$. An engineer needs two different numbers from this model and keeps confusing them.
Given
$V(t)=200-8t+0.08t^{2}$ litres, $0\le t\le 50$
Find
(a) Find the average rate of change of the volume over $[10,20]$, with units.
(b) Find the instantaneous rate of change at $t=10$ from the difference quotient, with units.
(c) Explain in one sentence why the two numbers differ in the direction they do.
Hint 1/4
Two different questions are being asked: one about an interval, one about an instant. Decide which formula belongs to which before computing.
Hint 2/4
Average over $[t_1,t_2]$ is $\frac{V(t_2)-V(t_1)}{t_2-t_1}$; the instantaneous rate at $a$ is $\lim_{h\to0}\frac{V(a+h)-V(a)}{h}$.
Hint 3/4
With $V(t)=200-8t+0.08t^{2}$: $V(10)=128$, $V(20)=72$, and $V(10+h)=128-6.4h+0.08h^{2}$.
Hint 4/4
So the average is $\frac{72-128}{10}=-5.6$ L/min and the instantaneous rate is $\lim_{h\to0}(-6.4+0.08h)=-6.4$ L/min.
the instant rate at the start of the interval is the faster drain, and the average includes later, slower minutes
Answer $$\boxed{-5.6\ \text{L/min average},\qquad -6.4\ \text{L/min at } t=10}$$
Check
Shrink the interval and watch the average approach the instantaneous value: over $[10,11]$ the average is $-6.32$ L/min, over $[10,10.1]$ it is $-6.392$ L/min.
The type was: difference quotient and limit, wearing an applied costume. The words average and instantaneous are the only signal, and they are worth a lot of marks.
Mistake ledger (15 entries)
⚠ Calling the average over a tiny interval the speed at the instant
the numbers get so close that the difference looks like rounding, and a calculator encourages it
Close the page and write down everything you can about limits from this section: what a limit is, the four things substitution can produce and what you do about each, and the three ways continuity fails. Then reopen and compare — the gaps you find are the list of what to reread, and nothing here is scored.
Turn an average rate over an interval into a rate at an instant, and explain why the direct attempt gives zero over zero?
c-instantaneous-rate
State what a limit ignores, and give a function whose limit and value at a point are different numbers?
c-limit-idea
Compute both one sided limits of a piecewise or absolute value function and apply the existence test without hesitating over which branch to use?
c-one-sided
Decide whether a zero of the denominator is a hole or an asymptote, and get the sign right on each side?
c-infinite-limits
Name the law behind each line of a computation, and say why a law fails when a denominator tends to zero?
c-limit-laws
Build a sandwich for an oscillating function and explain why the product law was unavailable?
c-squeeze
Run the three part continuity test at a point and classify the failure as removable, jump or infinite?
c-continuity
Glossary (15 terms)
limitlimit
The value the outputs of a function close in on as the input closes in on a point, computed without ever using the value at that point.
one sided limittek taraflı limit
The value approached when the input is restricted to one side of the point, written with a minus or plus superscript.
two sided limitiki taraflı limit
The ordinary limit, which exists exactly when both one sided limits exist and agree.
indeterminate formbelirsiz form
A result of substitution such as zero over zero that carries no value and instructs you to rewrite the expression.
infinite limitsonsuz limit
The case where outputs pass every bound near the point; a description of how the limit fails rather than a value.
vertical asymptotedüşey asimptot
A line x equal to a that the graph runs along without bound on at least one side.
kesen doğru
The straight line through two points of a curve, whose slope is an average rate of change.
tangent lineteğet doğru
The line whose slope is the limit of the secant slopes as the second point slides into the first.
difference quotientfark oranı
The quotient of the change in the output by the change in the input, the object every instantaneous rate is built from.
instantaneous rate of changeanlık değişim oranı
The limit of the average rates over shrinking intervals containing the instant.
Squeeze Theoremsıkıştırma teoremi
If a function is trapped between two functions with the same limit at a point, it has that limit too.
continuity at a pointbir noktada süreklilik
The property that the value approached and the value attained at the point are the same number.
removable discontinuitykaldırılabilir süreksizlik
A break where the limit exists but the value is missing or different, repairable by redefining that one value.
sıçrama süreksizliği
A break where both one sided limits are finite and different, which no single value can repair.
conjugateeşlenik
The expression obtained by flipping the sign between a root and the term next to it, used to clear roots from a quotient.
What comes next
§02 · Continuity, asymptotes, and limits at infinity
The two sides of a vertical asymptote told us what happens as x closes in on a fixed point. Next we let x itself run off to infinity, which turns the same machinery into horizontal asymptotes — and the three part continuity test grows into a theorem that locates solutions of equations nobody can solve.
Sources
James Stewart, Calculus (Metric Version), 9th edition — sections 1.4, 1.5, 1.6 and 1.8 Definitions, notation and the standard results follow this text, which is the required book for the course.
MATH 101 course syllabus: weekly plan and assessment weights Weeks 1 and 2 of the plan, and the grading weights quoted in the summary panel.
Conventions for reporting infinite limits and non existence An infinite limit is reported as a signed infinity when both sides agree, and as a pair of one sided statements when they do not.