Selecting Procedures for Determining Limits

AP Calculus · Limits and Continuity

Selecting Procedures for Determining Limits

No new technique here — only the skill of picking the right one from the shape of the expression.

key idea$\text{Substitute first. The form you get tells you the route.}$

How to choose

This topic teaches no new limit technique. What is new — and what the exam actually tests — is choosing the right technique fast, from the shape of the expression. So the work below is deliberately not a list of computations: it is a routing drill.

Step	What you do	Why it comes first
1	Substitute the value directly.	It resolves most limits outright and, when it fails, the way it fails names your route.
2	Read the form you got.	$\tfrac00$, $\tfrac{k}{0}$, $\tfrac{\infty}{\infty}$ and a plain number are four different situations.
3	Route on that form.	Only now do you factor, rationalise, use a known limit, or check one-sided behaviour.

Students lose time by manipulating before substituting. Substitution is diagnostic, not just a first attempt.

The decision tree

Substitute. Then find your case below.

CASE 1

You get a real number.That number is the limit — the function is continuous there. Stop. Do not simplify further.

CASE 2

You get $\tfrac{0}{0}$ — indeterminate.Something cancels. Route by what the expression looks like: polynomial → factor and cancel; a square root → multiply by the conjugate; compound fractions → clear the inner denominators; trig near $0$ → reshape into $\tfrac{\sin u}{u}$ or $\tfrac{1-\cos u}{u}$.

CASE 3

You get $\tfrac{k}{0}$ with $k\neq 0$.This is not indeterminate and nothing will cancel. The magnitude blows up — check the sign from each side. Equal signs → $+\infty$ or $-\infty$; opposite signs → the two-sided limit does not exist.

CASE 4

$x\to\pm\infty$ in a rational function.Compare degrees. Top < bottom → limit $0$. Top $=$ bottom → ratio of leading coefficients. Top > bottom → $\pm\infty$. (Or divide through by the highest power in the denominator.)

CASE 5

The function is piecewise, or has $|x|$, or a jump at the point.Substitution is meaningless until you pick a side. Compute the left and right limits separately; they must agree for the two-sided limit to exist.

CASE 6

It is a known special limit or a squeeze.$\lim_{x\to0}\tfrac{\sin x}{x}=1$ and $\lim_{x\to0}\tfrac{1-\cos x}{x}=0$ are facts you may quote. A bounded factor times something → $0$ is a squeeze, not an algebra problem.

Substituting gives…	Indeterminate?	Route	Typical finish
a real number $L$	no	done	$L$
$\tfrac00$, polynomials	yes	factor & cancel	substitute again
$\tfrac00$, with a radical	yes	conjugate	substitute again
$\tfrac00$, trig at $0$	yes	known limit / rewrite	$1$, $0$, or a multiple
$\tfrac{k}{0}$, $k\neq0$	no	one-sided signs	$\pm\infty$ or DNE
$\tfrac{\infty}{\infty}$	yes	compare degrees	$0$, ratio, or $\pm\infty$
different from each side	no	split the sides	DNE

The second column is the one students skip — and $\tfrac{k}{0}$ is where they skip it.

Worked examples

EX 1

Three limits that look alike, three different routes

routing

Evaluate $\lim\limits_{x\to2}\dfrac{x^{2}-4}{x-2}$, $\lim\limits_{x\to2}\dfrac{x^{2}+4}{x-2}$, and $\lim\limits_{x\to2}\dfrac{x^{2}-4}{x+2}$.

First one — substitute

$\tfrac{4-4}{2-2}=\tfrac00$. Indeterminate → CASE 2. Factor: $\tfrac{(x-2)(x+2)}{x-2}=x+2\to 4$.

Second one — substitute

$\tfrac{4+4}{0}=\tfrac{8}{0}$. Not indeterminate → CASE 3. Nothing cancels.

Finish the second by sides

As $x\to2^{+}$, $x-2\to0^{+}$ so the quotient $\to+\infty$; as $x\to2^{-}$ it $\to-\infty$. The two-sided limit does not exist.

Third one — substitute

$\tfrac{4-4}{2+2}=\tfrac{0}{4}=0$. A real number → CASE 1. Done, and factoring would have been wasted effort.

The lesson

The three numerators/denominators differ by a sign or a term, and each sends you down a different branch. Substituting first is what tells them apart.

EX 2

A radical — conjugate route

CASE 2

Evaluate $\lim\limits_{x\to0}\dfrac{\sqrt{x+9}-3}{x}$.

Substitute

$\tfrac{\sqrt9-3}{0}=\tfrac{0}{0}$ → CASE 2, radical present.

Multiply by the conjugate

$\dfrac{\sqrt{x+9}-3}{x}\cdot\dfrac{\sqrt{x+9}+3}{\sqrt{x+9}+3}=\dfrac{(x+9)-9}{x\left(\sqrt{x+9}+3\right)}$.

Cancel

$=\dfrac{x}{x\left(\sqrt{x+9}+3\right)}=\dfrac{1}{\sqrt{x+9}+3}$.

Substitute again

$\dfrac{1}{3+3}=\dfrac16$.

EX 3

Do not manipulate a $\tfrac{k}{0}$ form

CASE 3

Evaluate $\lim\limits_{x\to1}\dfrac{x+3}{(x-1)^{2}}$.

Substitute

$\tfrac{4}{0}$. Not indeterminate — no algebra will help.

Check the sign of the denominator

$(x-1)^{2}$ is a square, so it is positive from both sides.

Conclude

Numerator $\to4\gt 0$, denominator $\to0^{+}$ from both sides, so the limit is $+\infty$ — the same from each side.

Contrast

Had the denominator been $(x-1)$ instead, the signs would differ and the answer would be DNE. The even power is what makes it $+\infty$.

EX 4

Route by the form you get, not by the trig

CASE 6

Evaluate $\lim\limits_{x\to0}\dfrac{\sin(5x)}{3x}$.

Substitute

$\tfrac{\sin 0}{0}=\tfrac00$ → CASE 2/6.

Reshape to the known form

The known limit is $\tfrac{\sin u}{u}\to1$. Here $u=5x$, so force a $5x$ underneath: $\dfrac{\sin(5x)}{3x}=\dfrac{5}{3}\cdot\dfrac{\sin(5x)}{5x}$.

Apply the fact

$\dfrac{\sin(5x)}{5x}\to1$ as $x\to0$.

Finish

The limit is $\tfrac53$.

EX 5

Pick a side before you substitute

CASE 5

For $f(x)=\dfrac{|x-4|}{x-4}$, find $\lim\limits_{x\to4}f(x)$.

Recognise the trap

Substitution gives $\tfrac00$, but nothing factors — the absolute value hides a sign change. CASE 5.

Right side

For $x\gt 4$, $|x-4|=x-4$, so $f(x)=1$. Hence $\lim_{x\to4^{+}}f(x)=1$.

Left side

For $x\lt 4$, $|x-4|=-(x-4)$, so $f(x)=-1$. Hence $\lim_{x\to4^{-}}f(x)=-1$.

Conclude

The one-sided limits disagree, so $\lim_{x\to4}f(x)$ does not exist — even though the $\tfrac00$ form suggested cancellation.

EX 6

At infinity, compare degrees

CASE 4

Evaluate $\lim\limits_{x\to\infty}\dfrac{3x^{2}-x}{5x^{2}+7}$ and $\lim\limits_{x\to\infty}\dfrac{3x}{5x^{2}+7}$.

First: equal degrees

Both top and bottom are degree $2$, so the limit is the ratio of leading coefficients: $\tfrac35$.

Second: bottom wins

Degree $1$ over degree $2$. The denominator grows faster, so the limit is $0$.

Show the work if asked

Divide through by $x^{2}$: $\dfrac{3/x}{5+7/x^{2}}\to\dfrac{0}{5}=0$.

Guard

This degree rule is for $x\to\pm\infty$ only. At a finite point it says nothing.

Watch out

$\tfrac{k}{0}$ is not $\tfrac00$

Only $\tfrac00$ (and $\tfrac{\infty}{\infty}$) are indeterminate. With a non-zero numerator over $0$, no factoring exists — go straight to one-sided signs.

Manipulating before substituting

Factoring a limit that substitution would have resolved wastes exam time and invites algebra slips. Substitute first, every time.

“DNE” versus $\infty$

If both sides blow up the same way, saying $+\infty$ is more informative than DNE and is what the reader wants. If the sides disagree, DNE is the correct answer.

Quoting $\tfrac{\sin x}{x}$ with the wrong argument

$\lim_{x\to0}\tfrac{\sin(5x)}{x}$ is $5$, not $1$. The fact only applies when the argument of $\sin$ and the denominator are identical.

Course map. Topic 1.7 · The CED assigns this topic no unique learning objective — it is a synthesis of 1.5 (limit laws) and 1.6 (algebraic manipulation), assessed as procedure selection.