AP Calculus AB · Lesson 3

Asymptotes and limits at infinity

Know Know it

An infinite limit describes outputs that grow without bound near an input. lim⁡x→a+f(x)=∞\lim_{x \to a^+} f(x) = \infty means f(x)f(x) passes every height as xx approaches aa from the right. The limit still does not exist as a number; the symbol says how it fails. If ff grows without bound from at least one side of aa, the line x=ax = a is a vertical asymptote.

For a rational function, a zero of the denominator that does not cancel with the numerator gives a vertical asymptote; one that cancels gives a hole. For the sign of a one-sided infinite limit, check the signs of the top and the bottom just to that side.

A limit at infinity, lim⁡x→∞f(x)=L\lim_{x \to \infty} f(x) = L (or x→−∞x \to -\infty), describes the ends of the graph, and y=Ly = L is then a horizontal asymptote. For a rational function, compare degrees: a lower degree on top gives 0; equal degrees give the ratio of the leading coefficients; a higher degree on top gives no horizontal asymptote.

The graph of y = \frac{2x}{x - 1}−6−4−22468−6−4−22468100xy(0, 0)
The graph of y=2xx−1y = \frac{2x}{x - 1} Two branches. The dashed vertical line x = 1 and the dashed horizontal line y = 2 are asymptotes. Left of x = 1, the curve passes through (0, 0), rises toward 2 as x goes far left, and falls without bound as x approaches 1. Right of x = 1, it comes down from very large heights and levels off toward 2 from above.

Worked example Easy

What is lim⁡x→∞15x2−x5x2+7\lim_{x \to \infty}\frac{15x^2 - x}{5x^2 + 7}?

  1. 00
  2. 33Answer
  3. 1515
  4. ∞\infty

How to solve it

  1. Both the top and the bottom have degree 2.
  2. Divide every term by x2x^2: 15−1x5+7x2\frac{15 - \frac{1}{x}}{5 + \frac{7}{x^2}}.
  3. As x→∞x \to \infty the small terms vanish, leaving 155=3\frac{15}{5} = 3.

Why each choice is right or wrong

  • A. Incorrect. 0 is the limit when the top has the lower degree, but here the top and bottom both have degree 2.
  • B. Correct. The degrees match, so for large xx the quotient behaves like 15x25x2=3\frac{15x^2}{5x^2} = 3.
  • C. Incorrect. 15 is only the top's leading coefficient; the bottom's leading coefficient 5 divides it.
  • D. Incorrect. ∞\infty would need the top's degree to be higher, but both degrees are 2, so the quotient levels off.

Worked example Medium

What is lim⁡x→3−2x+5x−3\lim_{x \to 3^-}\frac{2x + 5}{x - 3}?

  1. −∞-\inftyAnswer
  2. ∞\infty
  3. 22
  4. 1111

How to solve it

  1. Substitution gives 110\frac{11}{0}: a nonzero number over 0, so the quotient is unbounded near 3.
  2. For xx slightly less than 3, the top is about 11 (positive) and the bottom is a small negative number.
  3. Positive over small negative is a large negative number, so the limit is −∞-\infty.

Why each choice is right or wrong

  • A. Correct. Near 3 the top is close to 11, a positive number. Just left of 3 the bottom x−3x - 3 is a tiny negative number, so the quotient is a huge negative number and heads to −∞-\infty.
  • B. Incorrect. ∞\infty is the right-hand limit; from the left, x−3x - 3 is negative, which makes the quotient negative.
  • C. Incorrect. 2 is the ratio of leading coefficients, which gives the limit as x→∞x \to \infty, not near x=3x = 3.
  • D. Incorrect. 11 is the value of the top at 3; dividing by a bottom that approaches 0 makes the quotient unbounded.

Explain Explain it

Task verbs: explain, determine

Dominant terms decide limits at infinity beyond rational functions too: exe^x eventually outgrows every power of xx, and every positive power of xx outgrows ln⁡x\ln x. Divide the top and bottom by the fastest-growing term and see what survives.

Watch roots: x2=∣x∣\sqrt{x^2} = |x|. For x<0x < 0 this is −x-x, so the sign can flip when x→−∞x \to -\infty.

Trap An asymptote can be crossed

A horizontal asymptote only describes the ends of a graph. The graph of sin⁡xx\frac{\sin x}{x} crosses its horizontal asymptote y=0y = 0 again and again while it approaches it. Do not rule out a horizontal asymptote just because the graph touches the line somewhere.

Worked example Hard

What is lim⁡x→−∞64x2+1x+2\lim_{x \to -\infty}\frac{\sqrt{64x^2 + 1}}{x + 2}?

  1. −8-8Answer
  2. 00
  3. 88
  4. 6464

How to solve it

  1. For x<0x < 0, x2=∣x∣=−x\sqrt{x^2} = |x| = -x, so 64x2+1=∣x∣64+1x2=−x64+1x2\sqrt{64x^2 + 1} = |x|\sqrt{64 + \frac{1}{x^2}} = -x\sqrt{64 + \frac{1}{x^2}}.
  2. Divide top and bottom by xx: the quotient is −64+1x21+2x\frac{-\sqrt{64 + \frac{1}{x^2}}}{1 + \frac{2}{x}}.
  3. As x→−∞x \to -\infty this approaches −81=−8\frac{-8}{1} = -8.

Why each choice is right or wrong

  • A. Correct. For large negative xx, 64x2+1\sqrt{64x^2 + 1} is about 8∣x∣=−8x8|x| = -8x, while x+2x + 2 is about xx, so the quotient is about −8xx=−8\frac{-8x}{x} = -8.
  • B. Incorrect. 0 would need the top to grow more slowly than the bottom, but both grow like a multiple of ∣x∣|x|.
  • C. Incorrect. 8 treats 64x2\sqrt{64x^2} as 8x8x, which is negative here; for x<0x < 0 it equals −8x-8x, a positive number.
  • D. Incorrect. 64 forgets the square root on 64x264x^2; the top grows like 8∣x∣8|x|, not like 64∣x∣64|x|.

Explain Your turn

Let r(x)=x2−4x2−3x+2r(x) = \frac{x^2 - 4}{x^2 - 3x + 2}. Both x=1x = 1 and x=2x = 2 make the denominator 0. Explain why the graph of rr has a vertical asymptote at x=1x = 1 but not at x=2x = 2.

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Compare with a model answer

A full answer does this

  • Shows that x−2x - 2 is a factor of both the top and the bottom, so it cancels
  • Says that near x=1x = 1 the top stays away from 0 while the bottom approaches 0, so rr is unbounded there
  • Says that rr approaches a finite number as xx approaches 2, so there is no asymptote at 2

Model answer

Factor: r(x)=(x−2)(x+2)(x−1)(x−2)r(x) = \frac{(x - 2)(x + 2)}{(x - 1)(x - 2)}. The factor x−2x - 2 is in both the top and the bottom and cancels, so r(x)=x+2x−1r(x) = \frac{x + 2}{x - 1} for x≠2x \ne 2. As xx approaches 2, r(x)r(x) approaches 41=4\frac{4}{1} = 4, a finite number, so the graph has a hole at (2,4)(2, 4), not an asymptote. As xx approaches 1, the top approaches 3 while the bottom approaches 0, so r(x)r(x) grows without bound and x=1x = 1 is a vertical asymptote.

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