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# 6.6: L’Hôpital’s Rule

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

A student will be able to:

• Learn how to find the limit of indeterminate form $(0/0)$ by L’Hospital’s rule.

If the two functions $f(x)$ and $g(x)$ are both equal to zero at $x = a,$ then the limit

$\lim_{x \to \ a} \frac {f(x)}{g(x)}$

cannot be found by directly substituting $x = a.$ The reason is because when we substitute $x = a,$ the substitution will produce $(0/0),$ known as an indeterminate form, which is a meaningless expression. To work around this problem, we use L’Hospital’s rule, which enables us to evaluate limits of indeterminate forms.

L’Hospital’s Rule

If $\lim_{x\to a}f(x) = \lim_{x\to a}g(x) = 0$, and $f'(a)$ and $g'(a)$ exist, where $g'(a) \neq 0$, then

$\lim_{x \to \ a} \frac {f(x)}{g(x)} = \lim_{x \to \ a} \frac {f'(x)}{g'(x)}.$

The essence of L’Hospital’s rule is to be able to replace one limit problem with a simpler one. In each of the examples below, we will employ the following three-step process:

1. Check that $\lim_{x \rightarrow a} \frac{f(x)} {g(x)}$ is an indeterminate form $0/0.$ To do so, directly substitute $x = a$ into $f(x)$ and $g(x).$ If you get $f(a) = g(a) = 0,$ then you can use L’Hospital’s rule. Otherwise, it cannot be used.
2. Differentiate $f(x)$ and $g(x)$ separately.
3. Find $\lim_{x \rightarrow a} \frac{f'(x)} {g'(x)}.$ If the limit is finite, then it is equal to the original limit $\lim_{x \rightarrow a} \frac{f(x)} {g(x)}$ .

Example 1:

Find $\lim_{x \rightarrow 0} \frac{\sqrt{2 + x} - \sqrt{2}} {x}.$

Solution:

When $x = 0$ is substituted, you will get $0/0.$

Therefore L’Hospital’s rule applies:

$\lim_{x \rightarrow 0} \frac{\sqrt{2 + x} - \sqrt{2}} {x} & = \lim_{x \rightarrow 0} \left [\frac{\frac{d} {dx} (\sqrt{2 + x } - \sqrt{2})} {\frac{d} {dx} (x)} \right]\\& = \left [\frac{1/(2\sqrt{2 + x})} {1} \right]_{x = 0}\\& = \frac{1} {2\sqrt{2}}\\& = \frac{\sqrt{2}} {4}.$

Example 2:

Find $\lim_{x \rightarrow 0} \frac{1 - \cos 2x} {x^2 + 2x}$

Solution:

We can see that the limit is $0/0$ when $x = 0$ is substituted.

Using L’Hospital’s rule,

$\lim_{x \rightarrow 0} \frac{1 - \cos 2x} {x^2 + 2x} & = \left [\frac{2 \sin 2x} {2x + 2} \right]_{x = 0}\\& = 0/2\\& = 0.$

Example 3:

Use L’Hospital’s rule to evaluate $\lim_{x \rightarrow 3} \frac{x^2 - 9} {x - 3}$ .

Solution:

$\lim_{x \rightarrow 3} = \frac{x^2 - 9} {x - 3} = \lim_{x \rightarrow 3} \frac{2x} {1} = 6.$

Example 4:

Evaluate $\lim_{x \rightarrow 0} \frac{\sin 3x} {x}.$

Solution:

$\lim_{x \rightarrow 0} \frac{\sin 3x} {x} = \lim_{x \rightarrow 0} \frac{3 \cos 3x} {1} = 3.$

Example 5:

Evaluate $\lim_{x \rightarrow \pi/2} \frac{5 - 5 \sin x} {\cos x}$ .

Solution:

We can use L’Hospital’s rule since the limit produces the $0/0$ once $x = \pi/2$ is substituted. Hence

$\lim_{x \rightarrow \pi/2} \frac{5 - 5 \sin x} {\cos x} = \lim_{x \rightarrow \pi/2} \frac{0 - 5 \cos x} {-\sin x} = \frac{0} {-1} = 0.$

A broader application of L’Hospital’s rule is when $x = a$ is substituted into the derivatives of the numerator and the denominator but both still equal zero. In this case, a second differentiation is necessary.

Example 6:

Evaluate $\lim_{x \rightarrow 0} \frac{e^x -\cos x} {x^2}.$

Solution:

$\lim_{x \rightarrow 0} \frac{e^x - \cos x} {x^2} = \lim_{x \rightarrow 0} \frac{e^x + \sin x} {2x}.$

As you can see, if we apply the limit at this stage the limit is still indeterminate. So we apply L’Hospital’s rule again:

$& = \lim_{x \rightarrow 0} \frac{e^x - \cos x} {2}\\& = \frac{1 - 1} {2} = \frac{0} {2} = 0.$

## Review Questions

Find the limits.

1. $\lim_{\theta \rightarrow 0} \frac{\tan \theta}{\theta}$
2. $\lim_{x \rightarrow 1} \frac{\ln x} {\tan \pi x}$
3. $\lim_{x \rightarrow 0} \frac{e^{10x} - e^{6x}} {x}$
4. $\lim_{x \rightarrow \pi} \frac{\sin x} {x - \pi}$
5. $\lim_{x \rightarrow 0} \frac{xe^x} {1 - e^x}$
6. If $k$ is a nonzero constant and $x > 0.$
1. Show that $\int_1^x \frac{1} {t^{1 - k}} dt = \frac{x^k - 1} {k}.$
2. Use L’Hospital’s rule to find $\lim_{k \rightarrow 0} \frac{x^k - 1} {k}.$
7. Cauchy’s Mean Value Theorem states that if the functions $f$ and $g$ are continuous on the interval $(a, b)$ and $g' \neq 0,$ then there exists a number $c$ such that $\frac{f'(c)} {g'(c)} = \frac{f(b) - f(a)} {g(b) - g(a)}.$ Find all possible values of $c$ in the interval $(a, b)$ that satisfy this property for $f(x) & = \cos x\\g(x) & = \sin x$ on the interval $[a, b] = \left [0, \frac{\pi} {2}\right ].$

1. $1$
2. $1/\pi$
3. $4$
4. $-1$
5. $-1$
6. $\ln x$
7. $\frac{\pi} {4}$

## Texas Instruments Resources

In the CK-12 Texas Instruments Calculus FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9731.

Feb 23, 2012

Dec 05, 2014