Unit 1 · Topic 1.14 Beta

Infinite Limits and Vertical Asymptotes

An infinite limit, lim as x → c, f(x) = ±∞ , says f grows without bound near c ; the limit does not exist as a number.

Practice 2: Connecting RepresentationsPractice 4: Communication and Notation

Question set for this topic

Part 1 · Hook

Why this matters

Squeeze the denominator of a fraction toward zero while the top stays put, and the fraction explodes. Lenses do this near their focal point: move an object toward it and the image size grows beyond every bound. Calculus has a compact way to say "grows beyond every bound" and to record which way it goes on each side.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Substituting x=4x = 4 into x+1x−4\frac{x + 1}{x - 4} gives 50\frac{5}{0}. Near 4, the expression

  1. is unbounded
  2. is close to 5
  3. is close to 0
  4. is close to 1
Show the answer

A nonzero number over a shrinking denominator grows without bound.

  • Correct: is unbounded:
  • is close to 5:
  • is close to 0:
  • is close to 1:

2. An infinite discontinuity at x=cx = c means the graph has

  1. a vertical asymptote x=cx = c
  2. a hole at x=cx = c
  3. a jump at x=cx = c
  4. a horizontal asymptote
Show the answer

An infinite discontinuity is unbounded behavior near c.

  • Correct: a vertical asymptote x=cx = c:
  • a hole at x=cx = c:
  • a jump at x=cx = c:
  • a horizontal asymptote:

Part 4 · See it

See it first

Two graphs with a dashed vertical asymptote at x = 2. Left: y = 1/(x − 2)², which rises without bound on both sides, so the limit as x approaches 2 is +∞ from each side. Right: y = 1/(x − 2), which falls without bound from the left (−∞) and rises without bound from the right (+∞), so the one-sided infinite limits have opposite signs.
At a vertical asymptote the two sides can go the same way or opposite ways. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. As x→cx \to c, the denominator shrinks to 0 while the numerator approaches a nonzero numberthe quotient grows without bound in size
  2. The sign of the quotient on each side of cc is found from the signs of its factorseach one-sided limit is ∞\infty or −∞-\infty
  3. At least one one-sided limit at cc is infinitethe line x=cx = c is a vertical asymptote
  4. ∞\infty is not a numberlimx→cf(x)=∞\lim_{x \to c} f(x) = \infty describes how the limit fails to exist; never write f(c)=∞f(c) = \infty

Part 6 · Key ideas

Key ideas

  • limx→cf(x)=∞\lim_{x \to c} f(x) = \infty means f(x)f(x) grows beyond every bound as xx nears cc. It is an infinite limit: the limit does not exist, and this says how.
  • Find each side by signs: a nonzero numerator over a denominator approaching 0 from the positive or negative side.
  • If the two sides agree, write limx→cf(x)=∞\lim_{x \to c} f(x) = \infty (or −∞-\infty). If they disagree, give each one-sided limit.
  • x=cx = c is a vertical asymptote when at least one one-sided limit at cc is ∞\infty or −∞-\infty.
  • For a rational function, cancel common factors first: a canceled factor gives a hole, a factor left in the denominator gives an asymptote.

Part 7 · Misconception

A common mistake

The wrong idea: limx→21(x−2)2\lim_{x \to 2} \frac{1}{(x - 2)^2}=∞= \infty, so the limit exists and its value is ∞\infty.

What actually happens: ∞\infty is not a number. An infinite limit is a limit that does not exist, described precisely: the values grow without bound. On a justification, say "unbounded" or "grows without bound", and never write f(2)=∞f(2) = \infty.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A rational function RR

The function R and its factored form
FormFormula
GivenR(x)R(x)=x+1x2−x−2= \frac{x + 1}{x^2 - x - 2}
FactoredR(x)R(x)=x+1(x−2)(x+1)= \frac{x + 1}{(x - 2)(x + 1)}

1. Which lines are vertical asymptotes of the graph of RR?

  1. The line x=2x = 2, and no other
  2. The lines x=2x = 2 and x=−1x = -1
  3. The line x=−1x = -1, and no other
  4. No line; RR has no vertical asymptote
Show the answer

The factor x+1x + 1 cancels, so R(x)=1x−2R(x) = \frac{1}{x - 2} for x≠−1x \ne -1: a hole at x=−1x = -1. The factor x−2x - 2 stays in the denominator, so RR is unbounded near 2.

  • Correct: The line x=2x = 2, and no other: x−2x - 2 remains in the denominator after cancelling; x+1x + 1 does not.
  • The lines x=2x = 2 and x=−1x = -1: At x=−1x = -1 the zero factor cancels, leaving a hole, not an asymptote.
  • The line x=−1x = -1, and no other: x=−1x = -1 is a hole. The asymptote is where the uncanceled factor x−2x - 2 is 0.
  • No line; RR has no vertical asymptote: Near x=2x = 2, R(x)=1x−2R(x) = \frac{1}{x - 2} grows without bound.

2. Find limx→2−R(x)\lim_{x \to 2^-} R(x).

  1. ∞\infty
  2. −∞-\infty
  3. 00
  4. −13-\frac{1}{3}
Show the answer

Near 2, R(x)=1x−2R(x) = \frac{1}{x - 2}. For xx slightly less than 2, x−2x - 2 is a tiny negative number, so 1x−2\frac{1}{x - 2} is a huge negative number: limx→2−R(x)=−∞\lim_{x \to 2^-} R(x) = -\infty.

  • ∞\infty: That is the right-hand limit, where x−2>0x - 2 > 0.
  • Correct: −∞-\infty: Positive 1 over a tiny negative number is very negative.
  • 00: The numerator is 1, not 0, while the denominator shrinks.
  • −13-\frac{1}{3}: −13-\frac{1}{3} is the height of the hole at x=−1x = -1.

3. Which statement about limx→2R(x)\lim_{x \to 2} R(x) is correct?

  1. limx→2R(x)=∞\lim_{x \to 2} R(x) = \infty, because RR is unbounded on both sides of 2
  2. limx→2R(x)=0\lim_{x \to 2} R(x) = 0, because 1x−2\frac{1}{x - 2} has a numerator of 1
  3. It does not exist; the sides go to −∞-\infty and ∞\infty
  4. It exists, because both one-sided limits are infinite in the same way
Show the answer

limx→2−R(x)=−∞\lim_{x \to 2^-} R(x) = -\infty and limx→2+R(x)=∞\lim_{x \to 2^+} R(x) = \infty. The sides head in opposite directions, so not even ∞\infty or −∞-\infty describes the two-sided behavior.

  • limx→2R(x)=∞\lim_{x \to 2} R(x) = \infty, because RR is unbounded on both sides of 2: Writing =∞= \infty for the two-sided limit needs both sides to go to +∞+\infty; here the left goes to −∞-\infty.
  • limx→2R(x)=0\lim_{x \to 2} R(x) = 0, because 1x−2\frac{1}{x - 2} has a numerator of 1: A fixed numerator over a shrinking denominator grows without bound.
  • Correct: It does not exist; the sides go to −∞-\infty and ∞\infty: Opposite signs on the two sides: describe each side separately.
  • It exists, because both one-sided limits are infinite in the same way: They are infinite in opposite directions; and an infinite limit is never a number anyway.

4. Find limx→3+x+13−x\lim_{x \to 3^+} \frac{x + 1}{3 - x}.

  1. ∞\infty
  2. −1-1
  3. −∞-\infty
  4. 44
Show the answer

As x→3+x \to 3^+, the numerator approaches 4 (positive) and 3−x3 - x is a tiny negative number. Positive over tiny negative: −∞-\infty.

  • ∞\infty: Check the sign of 3−x3 - x for xx just above 3: it is negative.
  • −1-1: −1-1 is the ratio of the xx coefficients, which describes behavior far from 3, not near it.
  • Correct: −∞-\infty: x>3x > 3 makes 3−x3 - x negative.
  • 44: 4 is the numerator's limit; the denominator heads to 0.

5. What are the vertical asymptotes of f(x)f(x)=x2−9x2−2x−3= \frac{x^2 - 9}{x^2 - 2x - 3}?

  1. x=3x = 3 and x=−1x = -1
  2. x=3x = 3, and no other line
  3. x=−1x = -1, and no other
  4. x=3x = 3 and x=−3x = -3
Show the answer

f(x)f(x)=(x−3)(x+3)(x−3)(x+1)= \frac{(x - 3)(x + 3)}{(x - 3)(x + 1)}=x+3x+1= \frac{x + 3}{x + 1} for x≠3x \ne 3. The factor x−3x - 3 cancels (a hole), and x+1x + 1 stays: one asymptote, x=−1x = -1.

  • x=3x = 3 and x=−1x = -1: At x=3x = 3 the factor cancels, so limx→3f(x)=32\lim_{x \to 3} f(x) = \frac{3}{2}: a hole.
  • x=3x = 3, and no other line: x=3x = 3 is the hole; the asymptote is x=−1x = -1.
  • Correct: x=−1x = -1, and no other: Only the uncanceled factor x+1x + 1 gives an asymptote.
  • x=3x = 3 and x=−3x = -3: x=−3x = -3 makes the numerator 0, which gives a zero of ff.

6. Which statement justifies that x=3x = 3 is a vertical asymptote of f(x)=5x−3f(x) = \frac{5}{x - 3}?

  1. f(3)f(3) is undefined
  2. The denominator of ff is 0 when x=3x = 3
  3. limx→3(x−3)\lim_{x \to 3} (x - 3)=0= 0
  4. limx→3+f(x)=∞\lim_{x \to 3^+} f(x) = \infty
Show the answer

A vertical asymptote x=cx = c is defined by an infinite one-sided limit at cc. Here limx→3+5x−3\lim_{x \to 3^+} \frac{5}{x - 3}=∞= \infty.

  • f(3)f(3) is undefined: Undefined points can be holes; that alone does not show an asymptote.
  • The denominator of ff is 0 when x=3x = 3: A zero denominator can come with a zero numerator and a hole; the limit must be shown infinite.
  • limx→3(x−3)\lim_{x \to 3} (x - 3)=0= 0: This is about the denominator only, not about the behavior of ff.
  • Correct: limx→3+f(x)=∞\lim_{x \to 3^+} f(x) = \infty: One infinite one-sided limit is exactly what a vertical asymptote means.

7. A student says: "limx→21(x−2)2\lim_{x \to 2} \frac{1}{(x - 2)^2}=∞= \infty, so this limit exists." What is the best response?

  1. It does not exist; =∞= \infty describes how the values grow
  2. It exists, because both sides of 2 agree on the value ∞\infty
  3. It exists, but just from the right side of x=2x = 2
  4. It does not exist, because the two sides go in opposite ways
Show the answer

∞\infty is not a real number, so an infinite limit is a limit that does not exist. Writing =∞= \infty adds information: the values grow without bound in the positive direction on both sides.

  • Correct: It does not exist; =∞= \infty describes how the values grow: "Equals infinity" is notation for a specific way of not existing.
  • It exists, because both sides of 2 agree on the value ∞\infty: Both sides do agree in direction, which is why we may write =∞= \infty; but ∞\infty is not a value.
  • It exists, but just from the right side of x=2x = 2: Both one-sided limits behave the same way; neither is a number.
  • It does not exist, because the two sides go in opposite ways: (x−2)2>0(x - 2)^2 > 0 on both sides, so both sides go up.

Part 9 · Summary

Summary

An infinite limit, limx→cf(x)=±∞\lim_{x \to c} f(x) = \pm\infty, says ff grows without bound near cc; the limit does not exist as a number. Find each side's sign from the signs of the numerator and the shrinking denominator. Any infinite one-sided limit at cc makes x=cx = c a vertical asymptote; a canceled factor makes a hole instead.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections