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Grade 12 Calculus · Lesson 1.590 min

Infinite Limits & Vertical Asymptotes

Read the side, predict the sign, factor before classifying, and translate infinite-limit notation into graph behavior.

Interpret unbounded one-sided behavior, find and classify vertical asymptotes, use sign analysis and multiplicity, and analyze rational and trigonometric functions.

Infinite behaviorGraphs & tablesVertical asymptotesFactor & classifySign analysisPower shortcutTrig VAsPropertiesSketching
01

Infinity describes behavior

What Does an Infinite Limit Mean?

Interactive lab

Read an infinite limit

Choose an approach side and output direction. Read the schematic, then match both parts of the limit notation.
x = cSchematic · not to scale
Which description matches?
Explanation

+∞+\infty and −∞-\infty are not real-number answers. They describe output growing without bound as the input approaches a value.

Worked example · Worked example · different rule

lim⁡x→4−−2x−4\lim_{x\to4^-}\frac{-2}{x-4} and lim⁡x→4+−2x−4\lim_{x\to4^+}\frac{-2}{x-4}

Predict the denominator sign on each side first.

02468-8-4048xy
Coordinate workspace — reveal the worked steps to compare the graph.

Predict before you reveal

For −2/(x−4)-2/(x-4), predict the left and right limits at 44, respectively.
Common mistake. lim⁡x→cf(x)=∞\lim_{x\to c}f(x)=\infty does not mean infinity is a real number.
02

Side first · direction second

Read Infinite Limits from Graphs and Tables

Explanation

For a graph: identify whether xx approaches from the left or right, then decide whether f(x)f(x) rises toward +∞+\infty or falls toward −∞-\infty.

Interactive lab

Infinite limits & asymptotes

Predict the direction, then move x toward 2. Compare both approach sides and powers. The original function is undefined at x = 2.
f(x)=1/(x−2)1f(x)=1/(x-2)^1y: ±80x = 24

Independent axes · y scale changes with the sample. Curves are clipped at the plot bounds, never joined across x = 2.

Optional Desmos view

The native graph and activity work without this external service.

x = 3 · x − 2 = 1

f(x) = 1

As x approaches 2 from the right, f(x) → ?
Explanation

Numerical lens

In a table, move inputs progressively closer to cc from one side at a time. Outputs with increasing positive magnitude indicate +∞+\infty; increasing negative magnitude indicates −∞-\infty.

Your turn

If lim⁡x→c−f(x)=−∞\lim_{x\to c^-}f(x)=-\infty and lim⁡x→c+f(x)=+∞\lim_{x\to c^+}f(x)=+\infty, what is true?
03

Unbounded from at least one side

Vertical Asymptotes

Definition

Vertical asymptote

The line x=cx=c is a vertical asymptote if the function approaches +∞+\infty or −∞-\infty as xx approaches cc from at least one side.
lim⁡x→c−f(x)=±∞orlim⁡x→c+f(x)=±∞⇒x=c is a VA\lim_{x\to c^-}f(x)=\pm\infty\quad\text{or}\quad\lim_{x\to c^+}f(x)=\pm\infty\Rightarrow x=c\text{ is a VA}

Worked example · Worked example · multiple candidates

f(x)=x+2x2−9f(x)=\frac{x+2}{x^2-9}

Factor the denominator and test whether either candidate cancels.

-6-3036-5-2.502.55xy
Coordinate workspace — reveal the worked steps to compare the graph.

Predict before you reveal

For (x+2)/(x2−9)(x+2)/(x^2-9), which excluded values remain vertical asymptotes?
04

Mr. Azzam’s routine

Factor · Classify · Sign-Check

Interactive lab

Factor before classifying

Factor the rule, retain its original exclusions, then decide whether each excluded input is a hole or a vertical asymptote.
f(x)=x2+2x−8x2−4f(x)=\frac{x^2+2x-8}{x^2-4}

Original domain: x ≠ −2, 2.

Try factoring · show a hint(x+4)(x−2)(x−2)(x+2)\frac{(x+4)(x-2)}{(x-2)(x+2)}

A hole requires a finite limit after simplification, not just a canceled factor.

Classify the original exclusions.

STEP 1

Factor

Factor numerator and denominator completely.

STEP 2

Candidates

Solve denominator = 0.

STEP 3

Classify

Completely cancelled denominator zero with finite reduced limit → hole. Remaining denominator zero → VA candidate.

STEP 4

Sign-check

Test left and right signs; then conclude.

Interactive lab

Signs determine the direction

Assume the numerator tends to a finite NONZERO number of the chosen sign, while the denominator tends to zero. Predict the quotient's infinite limit.

Numerator limit sign

Denominator approaches

(+)0+→  ?\frac{(+)}{0^+}\to\;?
What is the infinite-limit direction?

Worked example · Worked example · classify and sign-check

f(x)=x2−4x+3x2−x−6f(x)=\frac{x^2-4x+3}{x^2-x-6}

Factor before deciding what happens at either excluded value.

-6-3036-5-2.502.55xy
Coordinate workspace — reveal the worked steps to compare the graph.

Predict before you reveal

For (x2−4x+3)/(x2−x−6)(x^2-4x+3)/(x^2-x-6), classify 33 and −2-2.

Your turn

After factoring a rational function, a factor (x−c)(x-c) cancels completely. What should you suspect at x=cx=c?
05

Multiplicity predicts branch behavior

Even / Odd Power Shortcut

Interactive lab

Multiplicity and one-sided limits

For a fixed positive numerator 1, compare the denominator signs on the two sides of c. Predict the ordered pair (left limit, right limit).
f(x)=1(x−c)2f(x)=\frac{1}{(x-c)^2}x = cyxc − 3c + 3
Try the sign on each side
Approachx − cPower
Left: x < cNegative2
Right: x > cPositive2

The numerator stays positive here. A fixed negative numerator would reverse both directions.

(Left limit, right limit) = ?
ExplanationA(x)(x−c)2nvs.A(x)(x−c)2n+1\frac{A(x)}{(x-c)^{2n}}\quad\text{vs.}\quad\frac{A(x)}{(x-c)^{2n+1}}

Even multiplicity keeps the denominator sign; odd multiplicity changes it. Always account for the numerator sign. This shortcut assumes a positive integer n and a numerator continuous and nonzero at the target; simplify common factors first.

Your turn

Near x=cx=c, how does an even power (x−c)2n(x-c)^{2n} behave?
06

Rewrite as a quotient

Trigonometric Vertical Asymptotes

Interactive lab

Find trigonometric asymptotes

Rewrite as a quotient. Find the denominator zeros, and verify that the numerator is nonzero there. Angles are in radians.
tan⁡x=sin⁡xcos⁡x\tan x=\frac{\sin x}{\cos x}

Where does cos(x) = 0?

Unit-circle hint

sin(x) = 0 at integer multiples of π; cos(x) = 0 at odd multiples of π/2.

Which inputs are excluded and are vertical asymptotes? (n ∈ ℤ)

Worked example · Worked example · scaled cotangent

f(x)=cot⁡(2x)f(x)=\cot(2x)

Rewrite cotangent and solve where its denominator is zero.

−π−π/20π/2π-6-3036xy
Coordinate workspace — reveal the worked steps to compare the graph. · x in radians

Predict before you reveal

Where are all vertical asymptotes of cot⁡(2x)\cot(2x)?

Your turn

To locate possible vertical asymptotes of tan⁡x\tan x, which denominator should you inspect?
07

Combine finite and unbounded behavior carefully

Properties of Infinite Limits

ExplanationSum / difference with a finite limitf(x)→+∞,  g(x)→L⇒f(x)±g(x)→+∞f(x)\to+\infty,\;g(x)\to L\Rightarrow f(x)\pm g(x)\to+\infty
ExplanationProductL>0⇒f(x)g(x)→+∞,L<0⇒f(x)g(x)→−∞L>0\Rightarrow f(x)g(x)\to+\infty,\qquad L<0\Rightarrow f(x)g(x)\to-\infty
ExplanationFinite over unboundedg(x)f(x)→0\frac{g(x)}{f(x)}\to0

Worked example · Worked example · property with tangent

lim⁡x→(π/2)−5tan⁡x\lim_{x\to(\pi/2)^-}5\tan x

Find the one-sided tangent behavior before applying the constant.

−π−π/20π/2π-15-7.507.515xy
Coordinate workspace — reveal the worked steps to compare the graph. · x in radians

Predict before you reveal

Predict lim⁡x→(π/2)−5tan⁡x\lim_{x\to(\pi/2)^-}5\tan x.
08

Translate notation back into a picture

Sketch from Limit Information

Worked examplelim⁡x→2−f(x)=−∞,lim⁡x→2+f(x)=+∞\lim_{x\to2^-}f(x)=-\infty,\qquad\lim_{x\to2^+}f(x)=+\infty

Draw the vertical line x=2x=2. The left branch falls without bound; the right branch rises without bound.

-2-10123456-6-4-20246x = 2xy
One possible sketch; the limits determine nearby behavior, not the entire curve.

Worked example · Worked example · different asymptote

lim⁡x→−1−g(x)=+∞,lim⁡x→−1+g(x)=+∞\lim_{x\to-1^-}g(x)=+\infty,\qquad\lim_{x\to-1^+}g(x)=+\infty

Translate each one-sided limit before sketching.

-5-4-3-2-10123-6-4-20246x = -1xy
Sketch workspace — use the one-sided limits to draw each branch.

Predict before you reveal

Both one-sided limits at −1-1 are +∞+\infty. What must the nearby sketch show?
Key takeaway

Final concept check

Not every rational function has a VA. Denominator = 0 does not automatically mean VA. One-sided limits can approach opposite infinities.

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