L'Hôpital's Rule at Infinity

The form ∞/∞

L'Hôpital's rule also applies when the numerator and the denominator both grow without bound. If

\lim f(x)

and

\lim g(x)

are each

\infty

-\infty

, and

f

g

meet the same conditions as for

\frac{0}{0}

, then

\lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)},

provided the limit on the right exists or is

\pm\infty

. Here the limits can be as

x \to a

, from one side, or as

x \to \pm\infty

. As

x \to \infty

, the conditions are that

f

and

g

are differentiable with

g'(x) \ne 0

on some interval

(c, \infty)

, and similarly on

(-\infty, c)

x \to -\infty

. The rule for

\frac{0}{0}

works as

x \to \pm\infty

too.

Using the rule at infinity

Example 1

Find (a)

\lim_{x \to \infty} \frac{x^3}{e^{2x}}

and (b)

\lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}

(a)

\lim_{x \to \infty} x^3 = \infty

and

\lim_{x \to \infty} e^{2x} = \infty

, so

\lim_{x \to \infty} \frac{x^3}{e^{2x}} = \lim_{x \to \infty} \frac{3x^2}{2e^{2x}}.

Again

\lim_{x \to \infty} 3x^2 = \infty

and

\lim_{x \to \infty} 2e^{2x} = \infty

, so the rule applies a second time, and then, with

\lim_{x \to \infty} 6x = \infty

and

\lim_{x \to \infty} 4e^{2x} = \infty

, a third:

\begin{aligned} \lim_{x \to \infty} \frac{3x^2}{2e^{2x}} &= \lim_{x \to \infty} \frac{6x}{4e^{2x}} \\[4pt] &= \lim_{x \to \infty} \frac{6}{8e^{2x}} = 0. \end{aligned}

Each use lowers the power by one and keeps the exponential.

(b)

\lim_{x \to \infty} \ln x = \infty

and

\lim_{x \to \infty} \sqrt{x} = \infty

, so

\begin{aligned} \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} &= \lim_{x \to \infty} \frac{1/x}{1/(2\sqrt{x})} \\[4pt] &= \lim_{x \to \infty} \frac{2}{\sqrt{x}} = 0. \end{aligned}

Example 2

Find

\lim_{x \to \infty} \frac{4x^2 - \ln x}{2x^2 + 5x}

The numerator is

x^2\left(4 - \frac{\ln x}{x^2}\right)

, and

\frac{\ln x}{x^2} \to 0

by the growth rates, so

\lim_{x \to \infty}\left(4x^2 - \ln x\right) = \infty

. Also

\lim_{x \to \infty}\left(2x^2 + 5x\right) = \infty

, so

\lim_{x \to \infty} \frac{4x^2 - \ln x}{2x^2 + 5x} = \lim_{x \to \infty} \frac{8x - 1/x}{4x + 5}.

Now

\lim_{x \to \infty}(8x - 1/x) = \infty

and

\lim_{x \to \infty}(4x + 5) = \infty

, so a second use gives

\lim_{x \to \infty} \frac{8 + 1/x^2}{4} = 2.

A function known only through its derivative

Example 3

A function

f

, differentiable for

x > 0

, has

\lim_{x \to \infty} f(x) = 0

and

f'(x) = -\frac{6}{x^3} + 2e^{-2x}

. Find

\lim_{x \to \infty} x^2 f(x)

Write

x^2 f(x) = \frac{f(x)}{1/x^2}

. Then

\lim_{x \to \infty} f(x) = 0

and

\lim_{x \to \infty} \frac{1}{x^2} = 0

, the form

\frac{0}{0}

at infinity, and the rule needs only

f'

\begin{aligned} \lim_{x \to \infty} \frac{f(x)}{1/x^2} &= \lim_{x \to \infty} \frac{-6/x^3 + 2e^{-2x}}{-2/x^3} \\[4pt] &= \lim_{x \to \infty} \left(3 - x^3e^{-2x}\right) = 3. \end{aligned}

The last step uses

x^3e^{-2x} = \frac{x^3}{e^{2x}} \to 0

, from Example 1.

When the rule gives no answer

Example 4

Find

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

Since

x + \sin x \ge x - 1

\lim_{x \to \infty}(x + \sin x) = \infty

, and

\lim_{x \to \infty} x = \infty

. The form is

\frac{\infty}{\infty}

, but the rule leads to

\lim_{x \to \infty} (1 + \cos x)

, which doesn't exist:

1 + \cos x

keeps swinging between

0

and

2

. The rule then says nothing about the original limit. Divide by

x

instead:

\frac{x + \sin x}{x} = 1 + \frac{\sin x}{x}.

Since

-\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}

for

x > 0

, the squeeze theorem gives

\frac{\sin x}{x} \to 0

, so the limit is

1

(x + sin x)/x (black) settles toward 1, while the derivative ratio 1 + cos x (amber) keeps swinging.

Long-run behavior

Example 5

(a) The concentration of salt in a tank is

C(t) = \frac{6e^{t/2} - 6}{e^{t/2} + 2}

grams per liter for

t \ge 0

minutes. Find the concentration it approaches in the long run. (b) The current in a circuit is

I(t) = t^2e^{-t}

amperes. Find

\lim_{t \to \infty} I(t)

(a)

\lim_{t \to \infty}\big(6e^{t/2} - 6\big) = \infty

and

\lim_{t \to \infty}\big(e^{t/2} + 2\big) = \infty

, so

\lim_{t \to \infty} C(t) = \lim_{t \to \infty} \frac{3e^{t/2}}{\frac{1}{2}e^{t/2}} = \lim_{t \to \infty} 6 = 6.

The rule hands back the same exponential on top and bottom, so it cancels. In the long run the concentration approaches

6

grams per liter.

(b) Write

I(t) = \frac{t^2}{e^t}

. Then

\lim_{t \to \infty} t^2 = \infty

and

\lim_{t \to \infty} e^t = \infty

, so the limit equals

\lim_{t \to \infty} \frac{2t}{e^t}

. Again

\lim_{t \to \infty} 2t = \infty

and

\lim_{t \to \infty} e^t = \infty

, so it equals

\lim_{t \to \infty} \frac{2}{e^t} = 0

. The current dies away to

0

amperes.

A product rewritten as a quotient

Example 6

Find (a)

\lim_{x \to 0^+} x\ln x

and (b)

\lim_{x \to 0^+} x^x

(a) One factor approaches

0

and the other

-\infty

, so the product has no obvious limit. Rewrite it as a quotient:

x\ln x = \frac{\ln x}{1/x}

, with

\lim_{x \to 0^+} \ln x = -\infty

and

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

. Then

\begin{aligned} \lim_{x \to 0^+} x\ln x &= \lim_{x \to 0^+} \frac{1/x}{-1/x^2} \\[4pt] &= \lim_{x \to 0^+} (-x) = 0. \end{aligned}

(b)

x^x = e^{x\ln x}

, and the exponent approaches

0

by (a). Since

e^u

is continuous,

\lim_{x \to 0^+} x^x = e^0 = 1