02Continuity, asymptotes, and limits at infinity

Write down the full definition of continuity at $a$ and count how many conditions the given sentence has actually supplied.

Continuity at $a$ needs three things: $f(a)$ defined, the two sided limit existing, and the two being equal. Two finite one sided limits deliver at most part of the second.

The sentence never says the two sides are equal, and never mentions $f(a)$ at all.

False — a jump has two finite one sided limits, and so does a removable break.

Read what the asymptote statement actually claims. Is it a claim about every $x$, or about the behaviour of $f(x)$ as $x$ grows?

$y=L$ is a horizontal asymptote when $\lim_{x\to\infty}f(x)=L$. A limit constrains the tail, and puts no restriction on any individual value.

$\dfrac{\sin x}{x}$ is squeezed between $-\tfrac1x$ and $\tfrac1x$, so its limit at $\infty$ is $0$, and it equals $0$ at every $x=k\pi$.

True: that function crosses its asymptote infinitely often.

Write out the hypothesis in full and compare it with what you were given. One of the two halves is in trouble.

The theorem needs a value $N$ strictly between $f(a)$ and $f(b)$. If the two endpoint values are equal, no such $N$ exists.

Here $f(-1)=f(3)=4$, so there is no number strictly between them, and the theorem never starts.

The theorem yields nothing here — which does not mean $f$ has no zero, only that this tool cannot produce one.

Neither point can be classified from the denominator alone. Ask, at each of the two points, whether the numerator vanishes as well.

Factor both parts and cancel. A factor that dies on both floors leaves a hole; a surviving zero of the denominator with a nonzero numerator gives an infinite break.

$x^{2}-5x+6=(x-2)(x-3)$ and $x^{2}-4=(x-2)(x+2)$, so $f(x)=\dfrac{x-3}{x+2}$ for $x\neq2$. At $-2$ the reduced numerator is $-5$.

At $2$: removable, repaired by $f(2):=-\tfrac14$. At $-2$: infinite, with $+\infty$ from the left and $-\infty$ from the right, so $x=-2$ is the only vertical asymptote.

Continuity at $2$ forces $k$ to be one specific number. Which number is it, before you compute anything?

$k$ must equal $\lim_{x\to2}f(x)$. Substitution gives $\tfrac00$ because of the root, and the standard repair for a root in a $\tfrac00$ is multiplying by the conjugate.

Multiply top and bottom by $\sqrt{x+7}+3$: the numerator becomes $(x+7)-9=x-2$, and the given data is $f(x)=\frac{\sqrt{x+7}-3}{x-2}$ with $f(2)=k$.

After cancelling $x-2$ the limit is $\dfrac{1}{\sqrt{9}+3}=\dfrac16$, so $k=\tfrac16$.

Split the problem in two: what does the inside approach, and is the outside continuous there? Do not touch the cosine yet.

If $\lim_{x\to a}g(x)=L$ and $f$ is continuous at $L$, then $\lim f(g(x))=f(L)$. Cosine is continuous at every real number, so the hypothesis is free.

Inside: $\dfrac{\pi(x^{2}-9)}{x^{2}-3x}=\dfrac{\pi(x-3)(x+3)}{x(x-3)}=\dfrac{\pi(x+3)}{x}$ for $x\neq3$, which tends to $\dfrac{6\pi}{3}=2\pi$.

So the limit is $\cos(2\pi)=1$.

For each part, decide what the dominant power is and what you will divide by. For (b), decide the sign of that power on the side you are approaching from.

Divide by the dominant power. Inside a root, $\sqrt{x^{2}}=\vert x\vert$, and $\vert x\vert=-x$ when $x<0$.

(a) Divide by $x^{2}$: $\dfrac{2-1/x^{2}}{3+1/x}$. (b) $\sqrt{x^{2}+3x}=\vert x\vert\sqrt{1+3/x}=-x\sqrt{1+3/x}$ for $x<0$; divide top and bottom by $x$.

(a) $\dfrac{2}{3}$. (b) $\dfrac{-\sqrt{1}}{2}=-\dfrac12$.

Two competing giants cancel and leave something small. To see what is left, turn the difference into a single fraction.

Multiply by $\dfrac{\sqrt{4x^{2}+5x}+2x}{\sqrt{4x^{2}+5x}+2x}$; the numerator collapses to $\left(4x^{2}+5x\right)-4x^{2}=5x$. Then divide by the dominant power.

$w(x)=\dfrac{5x}{\sqrt{4x^{2}+5x}+2x}$, and here $x>0$, so $\sqrt{4x^{2}+5x}=x\sqrt{4+5/x}$.

Dividing by $x$: $\dfrac{5}{\sqrt{4+5/x}+2}\to\dfrac{5}{2+2}=\dfrac54$.

You are not looking for large changes in value; you are looking for one specific event between consecutive samples.

A zero is forced between two consecutive samples exactly when their signs differ, because then $N=0$ lies strictly between the two values and the theorem applies on that subinterval.

The signs of $-3,2,1,-4,-1,6$ are $-,+,+,-,-,+$, so the sign changes between the 1st and 2nd samples, the 3rd and 4th, and the 5th and 6th.

The guaranteed intervals are $(0,1)$, $(2,3)$ and $(4,5)$.

Two separate questions in one: what happens at the points where the formula refuses, and what happens at the two ends. Do them in that order.

Factor the denominator and check whether the numerator shares any zero. Then compare degrees: equal degrees give the ratio of leading coefficients.

$x^{2}-x-6=(x-3)(x+2)$, and $3x^{2}+5$ is $32$ at $x=3$ and $17$ at $x=-2$, so nothing cancels. Degrees are $2$ and $2$, leading coefficients $3$ and $1$.

So the vertical asymptotes are $x=3$ and $x=-2$, and the horizontal asymptote is $y=3$.

Ask what has to be true of the numerator at $x=2$ for the quotient to have a finite limit there, given that the denominator vanishes.

A finite limit at a zero of the denominator requires the numerator to vanish there too; otherwise the quotient is nonzero over zero, which is an infinite break. Then continuity forces $b=\lim_{x\to2}f(x)$.

Numerator at $2$: $4+2a-6=2a-2$. Setting it to $0$ gives $a=1$, and then $x^{2}+x-6=(x-2)(x+3)$.

So $a=1$, the limit is $2+3=5$, hence $b=5$; and for $a\neq1$ the break is infinite.

Try substitution first and name precisely what goes wrong. That name decides which repair is available.

Substitution gives $\tfrac00$, and the numerator is a difference of two roots, so multiply by the conjugate $\sqrt{1+x}+\sqrt{1-x}$.

The numerator becomes $(1+x)-(1-x)=2x$, so $m(x)=\dfrac{2x}{x\left(\sqrt{1+x}+\sqrt{1-x}\right)}$ for $x\neq0$.

Cancelling $x$ and substituting: $\dfrac{2}{1+1}=1$.

The numerator does not settle anywhere, so no limit law applies to it. Ask instead what the numerator can never do.

$-1\le\sin x\le1$ for every $x$; dividing by a positive $x$ preserves the inequalities, and the Squeeze Theorem finishes the job.

For $x>0$: $-\dfrac{1}{x}\le\dfrac{\sin x}{x}\le\dfrac{1}{x}$, and both bounds tend to $0$ as $x\to\infty$.

So the limit is $0$, the asymptote is $y=0$, and the graph meets it at every $x=k\pi$ with $k=1,2,3,\dots$ — infinitely often.

Each side of the junction has its own formula; decide which one is valid on which side before computing anything.

The left limit uses the piece valid for $x<1$, the right limit the piece valid for $x>1$, and the value $f(1)$ comes from the piece whose inequality contains $1$. An infinite one sided limit makes the break infinite and the line $x=1$ an asymptote.

Left of $1$ the formula is $x^{2}$, so the left limit is $1$, and $f(1)=1^{2}=1$ because $x\le1$ owns the junction. Right of $1$ the formula is $\frac{1}{x-1}$, whose denominator shrinks to $0$ through positive values.

So the left limit is $1$, the right limit is $+\infty$, the break is infinite with $x=1$ a vertical asymptote, and $f$ is continuous from the left because $f(1)=1$ equals the left limit.