Infinite Limits and Asymptotes

Take a quotient of two functions continuous at

a

, with the denominator nonzero near

a

except at

a

itself. If substituting

a

gives

\frac{b}{0}

with

b \ne 0

, the size of the quotient grows without bound near

a

, so

x = a

is a vertical asymptote. Its sign on each side comes from the sign of the numerator and the sign the denominator has on that side, a method called sign analysis. The two sides can have the same sign or opposite signs.

Example 1

Find each one-sided limit.

\begin{aligned} &\text{(a)}\;\; \lim_{x \to 2^-} \frac{x + 1}{x^2 - 4}, \\[4pt] &\phantom{\text{(a)}\;\;} \lim_{x \to 2^+} \frac{x + 1}{x^2 - 4} \\[12pt] &\text{(b)}\;\; \lim_{x \to (\pi/2)^-} \tan x, \;\; \lim_{x \to (\pi/2)^+} \tan x \\[12pt] &\text{(c)}\;\; \lim_{x \to 1^-} \frac{1}{\ln x}, \;\; \lim_{x \to 1^+} \frac{1}{\ln x} \\[12pt] &\text{(d)}\;\; \lim_{x \to (-3)^-} \frac{2 - x}{(x + 3)^2}, \\[4pt] &\phantom{\text{(d)}\;\;} \lim_{x \to (-3)^+} \frac{2 - x}{(x + 3)^2} \end{aligned}

(a) The numerator approaches

3

. The denominator is

(x - 2)(x + 2)

. Just left of

2

x - 2

is negative and

x + 2

is positive, so the denominator approaches

0

through negative values, and the quotient approaches

-\infty

. Just right of

2

, the denominator is positive, and the quotient approaches

\infty

y = (x + 1)/(x² − 4), with vertical asymptotes x = −2 and x = 2.

(b) Write

\tan x = \frac{\sin x}{\cos x}

. Near

\frac{\pi}{2}

the numerator approaches

1

. The cosine is positive just left of

\frac{\pi}{2}

and negative just right of it, so the limits are

\infty

from the left and

-\infty

from the right.

(c) Here

\ln x

approaches

0

, negative for

x < 1

and positive for

x > 1

. So

\frac{1}{\ln x}

approaches

-\infty

from the left and

\infty

from the right.

(d) The numerator approaches

5

, and the denominator

(x + 3)^2

is positive on both sides of

-3

. Both one-sided limits are

\infty

, so

\lim_{x \to -3} \frac{2 - x}{(x + 3)^2} = \infty

Limits at infinity

The basic limit is

\lim_{x \to \pm\infty} \frac{1}{x^n} = 0

for every positive integer

n

. To use it on a rational function, divide the numerator and the denominator by the highest power of

x

in the denominator, which gives three cases.

Example 2

Find each limit.

\begin{aligned} &\text{(a)}\;\; \lim_{x \to \infty} \frac{6x^2 - x}{2x^2 + 5} \\[10pt] &\text{(b)}\;\; \lim_{x \to -\infty} \frac{x^3 - 2x}{4x^2 + 1} \\[10pt] &\text{(c)}\;\; \lim_{x \to \infty} \frac{\sqrt{4x^2 + 1}}{x - 3}, \\[4pt] &\phantom{\text{(c)}\;\;} \lim_{x \to -\infty} \frac{\sqrt{4x^2 + 1}}{x - 3} \end{aligned}

(a) Divide the numerator and the denominator by

x^2

\begin{aligned} \lim_{x \to \infty} \frac{6x^2 - x}{2x^2 + 5} &= \lim_{x \to \infty} \frac{6 - \frac{1}{x}}{2 + \frac{5}{x^2}} \\[4pt] &= \frac{6}{2} = 3. \end{aligned}

(b) Divide by

x^2

. The numerator becomes

x - \frac{2}{x}

, which approaches

-\infty

, and the denominator approaches

4

. So the limit is

-\infty

(c) For

x \ne 0

\sqrt{4x^2 + 1} = |x|\sqrt{4 + \frac{1}{x^2}}

. For

x > 0

|x| = x

, and dividing above and below by

x

gives

\frac{\sqrt{4 + \frac{1}{x^2}}}{1 - \frac{3}{x}} \to \frac{2}{1} = 2.

For

x < 0

|x| = -x

, so the same division leaves a minus sign, and the limit is

-2

. The graph has two horizontal asymptotes,

y = 2

on the right and

y = -2

on the left.

y = √(4x² + 1)/(x − 3): horizontal asymptotes y = 2 and y = −2, approached slowly, and the vertical asymptote x = 3.

Comparing growth

For x > 0, 2ˣ and x² cross at x = 2 and at x = 4, and past 4, 2ˣ pulls away. The graph of ln x stays below both and barely rises.

Example 4

Find

\lim_{x \to \infty} \frac{x + \cos x}{x}

, and describe how the graph of

y = \frac{x + \cos x}{x}

meets its horizontal asymptote.

For

x > 0

, split the fraction:

\frac{x + \cos x}{x} = 1 + \frac{\cos x}{x}

. Since

-1 \le \cos x \le 1

1 - \frac{1}{x} \le \frac{x + \cos x}{x} \le 1 + \frac{1}{x}.

Both bounds approach

1

x \to \infty

. The squeeze theorem works the same way at infinity, with the inequalities holding for all large

x

. So the limit is

1

, and

y = 1

is a horizontal asymptote. For

x > 0

, the graph crosses it wherever

\cos x = 0

, at

x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots

, infinitely often. A horizontal asymptote describes the graph far out, and crossing it is allowed.

y = (x + cos x)/x between the dashed graphs y = 1 ± 1/x, crossing y = 1 at every odd multiple of π/2.

Example 5

Find every vertical and horizontal asymptote of

f(x) = \frac{2x^2 - 2}{x^2 - x - 6},

with the one-sided behavior at each vertical asymptote.

Factor:

f(x) = \frac{2(x - 1)(x + 1)}{(x - 3)(x + 2)}

. No factor cancels. The denominator is

0

x = 3

and

x = -2

, and the numerator isn't, so both are vertical asymptotes. Near

3

the numerator is near

16

, and

(x - 3)(x + 2)

is negative on the left and positive on the right:

\begin{aligned} \lim_{x \to 3^-} f(x) &= -\infty, \\[4pt] \lim_{x \to 3^+} f(x) &= \infty. \end{aligned}

Near

-2

the numerator is near

6

, and

(x - 3)(x + 2)

is positive on the left and negative on the right:

\begin{aligned} \lim_{x \to (-2)^-} f(x) &= \infty, \\[4pt] \lim_{x \to (-2)^+} f(x) &= -\infty. \end{aligned}

Dividing above and below by

x^2

\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{2 - \frac{2}{x^2}}{1 - \frac{1}{x} - \frac{6}{x^2}} = 2.

The limit as

x \to -\infty

is also

2

, so

y = 2

is the horizontal asymptote. Setting

f(x) = 2

gives

2x^2 - 2 = 2x^2 - 2x - 12

, so

x = -5

. The graph crosses its horizontal asymptote at

(-5, 2)

The graph of f: asymptotes x = −2, x = 3, and y = 2.

Zoomed in near (−5, 2), with horizontal gridlines 0.1 apart: the graph crosses y = 2 from below.