3.6 L’Hopital’s Rule

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In this section we compute limits using L’Hopital’s Rule which requires our knowledge of derivatives.

L’Hopital’s Rule

L’Hopital’s Rule uses the derivative to help us find limits involving indeterminate forms. The main indeterminate forms we will discuss are $\frac 00, \frac {\infty }{\infty }, 0\cdot \infty , 1^\infty$ and $0^0$ . We begin with the fractional forms.

L’Hopital’s Rule

$\text {If} \;\; \lim _{x \to c} \frac {f(x)}{g(x)} = \frac {0}{0} \; \text {or} \; \frac {\pm \infty }{\pm \infty }, \;\;\text {then} \; \lim _{x \to c} \frac {f(x)}{g(x)} = \lim _{x \to c} \frac {f'(x)}{g'(x)}$ provided the latter limit exists. The statement is also true for one-sided limits and if ‘ $c$ ’ is replaced by $\infty$ or $-\infty$ .

The indeterminate form $\frac 00$

example 1 Compute the limit: $\displaystyle {\;\lim _{x \to 0} \frac {\sin (3x)}{6x}}.$

Plugging in the terminal value, $x=0$ , yields the indeterminate form $\frac 00$ , so L’Hopital’s rule applies.

We have

$\lim _{x \to 0} \frac {\sin (3x)}{6x} = \lim _{x \to 0} \frac {3\cos (3x)}{6} = \frac 12$

(problem 1a) Compute $\lim _{x \to 0} \frac {1 - \cos (x)}{x} = \answer {0}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

(problem 1b) Compute $\lim _{x \to 0} \frac {\tan (x)}{3x} = \answer {1/3}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

The derivative of $\tan (x)$ is $\sec ^2(x)$

Compute the limit of the new fraction

(problem 1c)

Let $m$ and $m$ be non-zero constants. Compute

$\lim _{x \to 0} \frac {\sin (mx)}{nx} = \answer {m/n}$

Check for a fractional indeterminate form.

Use the chain rule in the numerator.

(problem 1d)

Compute

$\lim _{x \to 0} \frac {\sin ^2(x)}{3x} = \answer {0}$

Check for a fractional indeterminate form.

Use the chain rule in the numerator.

example 2 Compute the limit: $\displaystyle {\;\lim _{x \to 0} \frac {e^{4x} - 1}{12x}}.$
Plugging in the terminal value, $x=0$ , yields the indeterminate form $\frac 00$ , so L’Hopital’s rule applies. We have $\lim _{x \to 0} \frac {e^{4x} - 1}{12x} = \lim _{x \to 0} \frac {4e^{4x}}{12} = \frac 13$

(problem 2a) Compute $\lim _{x \to 0} \frac {5x}{2 - 2e^x} = \answer {-5/2}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

(problem 2b) Compute $\lim _{x \to 0} \frac {e^{2x} -1}{7\sin (x)} = \answer {2/7}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

The derivative of $e^{2x}$ is $2e^{2x}$ , by the Chain Rule

Compute the limit of the new fraction

(problem 2c) Let $a$ and $b$ be non-zero constants. Compute $\lim _{x \to 0} \frac {e^{ax} - 1}{bx} = \answer {a/b}$

Check for fractional indeterminate form.

Use the chain rule in the numerator.

example 3 Compute $\lim _{x \to 2} \frac {x^3 - 8}{x^5 - 32}.$

Plugging in the terminal value, $x=2$ , gives the indeterminate form $\frac 00$ , so we can use L’Hopital’s Rule:

$\lim _{x \to 2} \frac {x^3 - 8}{x^5 - 32} = \lim _{x \to 2} \frac {3x^2}{5x^4} = \frac {12}{80} = \frac {3}{20}.$

(problem 3a) Compute $\lim _{x \to 3} \frac {x^3 - 27}{x^4 - 81} = \answer {1/4}$

Plug in $x=3$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

(problem 3b) Compute $\lim _{x \to 4} \frac {x^2 - 3x - 4}{x^3 - 64} = \answer {5/48}$

Plug in $x=4$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

(problem 3c) Let $m$ and $n$ be non-zero constants. Compute $\lim _{x \to a} \frac {x^m - a^m}{x^n - a^n} = \answer {m/n a^{m-n}}$

Check for fractional indeterminate form.

Use the power rule in the numerator and denominator.

The derivative of a constant is zero.

Sometimes we have to use L’Hopital’s Rule more than once.

example 4 Compute the limit: $\displaystyle {\;\lim _{x \to 0} \frac {e^x - x - 1}{x^2}}.$

Plugging in the terminal value, $x=0$ , yields the indeterminate form $\frac 00$ , so L’Hopital’s rule applies. We have $\lim _{x \to 0} \frac {e^x - x - 1}{x^2} = \lim _{x \to 0} \frac {e^x -1}{2x} = \frac {0}{0}.$ Applying L’Hopital’s Rule again gives $\lim _{x \to 0} \frac {e^x -1}{2x} = \frac {e^x}{2} = \frac {1}{2}.$ Hence $\displaystyle {\lim _{x \to 0} \frac {e^x - x - 1}{x^2} = \frac 12}.$

(problem 4a) Compute $\lim _{x \to 0} \frac {1 - \cos (x)}{x^2} = \answer {1/2}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

If you get $\frac 00$ , use L’Hopital’s Rule again

(problem 4b) Compute $\lim _{x \to 0} \frac {\sec (x) - 1}{x^3 - x^2} = \answer {-1/2}$

Plug in $x=0$

If you got $\frac 00$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

The derivative of $\sec (x)$ is $\sec (x)\tan (x)$

Compute the limit of the new fraction

If you get $\frac 00$ , use L’Hopital’s Rule again

Use the Product Rule to find the derivative of $\sec (x)\tan (x)$

example 5 Compute the limit: $\displaystyle {\;\lim _{x \to 0} \frac {\tan (x)-x}{x^3}}.$
Plugging in the terminal value, $x=0$ , yields the indeterminate form $\frac 00$ , so L’Hopital’s rule applies. We have $\lim _{x \to 0} \frac {\tan (x)-x}{x^3} = \lim _{x \to 0} \frac {\sec ^2(x)-1}{3x^2} = \frac {0}{0}.$ Applying L’Hopital’s Rule again gives $\lim _{x \to 0} \frac {\sec ^2(x)-1}{3x^2} = \lim _{x \to 0} \frac {2\sec ^2(x)\tan (x)}{6x} = \frac {0}{0}.$ We need to apply L’Hopital’s Rule again, but first, the numerator is complicated and $\lim _{x \to 0} \sec (x) = 1 \neq 0$ so we take a simplifying step before applying the rule. $\lim _{x \to 0} \frac {2\sec ^2(x)\tan (x)}{6x} = \lim _{x \to 0} \sec ^2(x) \cdot \lim _{x \to 0} \frac {2\tan (x)}{6x}=$ $1\cdot \lim _{x \to 0} \frac {2\sec ^2(x)}{6} =\frac {1}{3}.$

We next consider problems of the form $\frac {\infty }{\infty }$ . These are handled the same way as the $\frac 00$ case above.

The $\dfrac {\infty }{\infty }$ case.

example 6 Compute the limit: $\displaystyle {\;\lim _{x \to \infty } \frac {x^2}{e^x}}.$

As $x$ approaches $\infty$ we get the indeterminate form $\frac {\infty }{\infty }$ so L’Hopital’s Rule applies. We have $\lim _{x\to \infty } \frac {x^2}{e^x} = \lim _{x \to \infty } \frac {2x}{e^x} = \frac {\infty }{\infty }.$ Applying L’Hopital again, we get $\lim _{x \to \infty } \frac {2x}{e^x} = \lim _{\infty } \frac {2}{e^x} = 0.$ Hence $\lim _{x\to \infty }\frac {x^2}{e^x}=0$ . This limit can be generalized as follows: $\lim _{x\to \infty }\frac {x^n}{e^x} =0$ for any exponent $n = 1, 2, 3, \dots$ . This general result comes from using L’Hopital’s Rule $n$ times, yielding $\lim _{x\to \infty } \frac {n!}{e^x} =0$ where $n! = n(n-1)(n-2)\cdot \dots \cdot 3\cdot 2\cdot 1$ . The interpretation of this limit is that the exponential function $e^x$ grows faster than any power of $x$ as $x \to \infty$ .

(problem 6a) Compute $\lim _{x \to \infty } \frac {3x+2}{2x + 3} = \answer {3/2}$

“Plug in” $x=\infty$

If you got $\frac {\infty }{\infty }$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

(problem 6b) Compute $\lim _{x \to \infty } \frac {5x^2 + 2x - 3}{4x^2 - x + 1} = \answer {5/4}$

“Plug in” $x=\infty$

If you got $\frac {\infty }{\infty }$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

Compute the limit of the new fraction

If you got $\frac {\infty }{\infty }$ , use L’Hopital’s Rule again

example 7 Compute the limit: $\displaystyle {\;\lim _{x \to \infty } \frac {\sqrt {x}}{\ln (x)}}$ . As $x \to \infty$ we get $\frac {\infty }{\infty }$ , so L’Hopital’s Rule applies. We have: $\lim _{x \to \infty } \frac {\sqrt x}{\ln (x)} = \lim _{x \to \infty }\frac {\left (\frac {1}{2\sqrt x}\right )}{\left (\frac {1}{x}\right )}$ which simplifies to $\lim _{x\to \infty } \frac {x}{2\sqrt x} = \lim _{x\to \infty }\frac {\sqrt x}{2} = \infty .$ Hence, $\lim _{x\to \infty } \frac {\sqrt x}{\ln (x)} = \infty$ . The interpretation of this limit is that $\sqrt x$ goes to $\infty$ faster than $\ln (x)$ as $x\to \infty$ .

(problem 7a) Compute $\lim _{x \to \infty } \frac {\sqrt x}{2x + 3} = \answer {0}$

“Plug in” $x=\infty$

If you got $\frac {\infty }{\infty }$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

The derivative of $\sqrt x$ is $\frac {1}{2\sqrt x}$

Compute the limit of the new fraction

(problem 7b) Compute $\lim _{x \to \infty } \frac {\ln (x)}{1 + x^2} = \answer {0}$

“Plug in” $x=\infty$

If you got $\frac {\infty }{\infty }$ , use L’Hopital’s Rule

Take the derivative of the numerator and denominator separately

The derivative of $\ln (x)$ is $\frac {1}{x}$

Simplify the new fraction and then compute the limit

The $0 \cdot \infty$ case

L’Hopital’s Rule requires a fractional indeterminate form such as $\frac 00$ or $\frac {\infty }{\infty }$ , but we can use it to handle other indeterminate forms by rewriting expressions as fractions.

Examples of the $0\cdot \infty$ case.

example 8 Compute the limit: $\displaystyle {\;\lim _{x \to 0^+}x^2 \ln (x)}$ .

As $x\to 0^+$ we get $0 \cdot (-\infty )$ which is an indeterminate form, but L’Hopital’s Rule does not apply in this situation. We must rewrite the problem as a fraction, in the following way: $\lim _{x \to 0^+}\frac {\ln (x)}{x^{-2}}.$ Notice that this is equivalent to the original problem since $x^2 = \frac {1}{x^{-2}}.$ Also note that $x^{-2} =\frac {1}{x^2} \to \infty$ as $x\to 0^+$ . Now, we can use L’Hopital’s Rule because $\lim _{x \to 0+}\frac {\ln (x)}{x^{-2}}= \frac {-\infty }{\infty }.$ We get $\lim _{x \to 0^+}\frac {\ln (x)}{x^{-2}}= \lim _{x\to 0^+} \frac {1/x}{-2x^{-3}}$ which simplifies to $\lim _{x \to 0^+}-\frac {x^2}{2}= 0.$ Hence, $\lim _{x \to 0^+} x^2 \ln (x)=\lim _{x \to 0^+}\frac {\ln (x)}{x^{-2}}= 0.$

(problem 8a) Compute $\lim _{x \to 0^+} x^3\ln (x) = \answer {0}$

(problem 8b) Compute $\lim _{x \to \infty } 3xe^{-x} = \answer {0}$

“Plug in” $x=\infty$

If you got $\infty \cdot 0$ , rewrite the expression as a fraction

Take advantage of negative exponents: $e^{-x} = \frac {1}{e^x}$

Here is a detailed, lecture style video on L’Hopital’s Rule:

L’Hopital’s Rule

The indeterminate form \frac 00

The \dfrac {\infty }{\infty } case.

The 0 \cdot \infty case

The indeterminate form $\frac 00$

The $\dfrac {\infty }{\infty }$ case.

The $0 \cdot \infty$ case