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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)\begin{align*}(0/0)\end{align*} by L’Hospital’s rule.

If the two functions f(x)\begin{align*}f(x)\end{align*} and g(x)\begin{align*}g(x)\end{align*} are both equal to zero at x=a,\begin{align*}x = a,\end{align*} then the limit

limx af(x)g(x)

cannot be found by directly substituting x=a.\begin{align*}x = a.\end{align*} The reason is because when we substitute x=a,\begin{align*}x = a,\end{align*} the substitution will produce (0/0),\begin{align*}(0/0),\end{align*} 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 limxaf(x)=limxag(x)=0\begin{align*} \lim_{x\to a}f(x) = \lim_{x\to a}g(x) = 0\end{align*}, and f(a)\begin{align*} f'(a)\end{align*} and g(a)\begin{align*} g'(a)\end{align*} exist, where g(a)0\begin{align*}g'(a) \neq 0\end{align*}, then

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

Example 1:

Find limx02+x2x.\begin{align*}\lim_{x \rightarrow 0} \frac{\sqrt{2 + x} - \sqrt{2}} {x}.\end{align*}

Solution:

When x=0\begin{align*}x = 0\end{align*} is substituted, you will get 0/0.\begin{align*}0/0.\end{align*}

Therefore L’Hospital’s rule applies:

limx02+x2x=limx0ddx(2+x2)ddx(x)=[1/(22+x)1]x=0=122=24.

Example 2:

Find limx01cos2xx2+2x\begin{align*} \lim_{x \rightarrow 0} \frac{1 - \cos 2x} {x^2 + 2x}\end{align*}

Solution:

We can see that the limit is 0/0\begin{align*}0/0\end{align*} when x=0\begin{align*}x = 0\end{align*} is substituted.

Using L’Hospital’s rule,

Example 3:

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

Solution:

Example 4:

Evaluate \begin{align*} \lim_{x \rightarrow 0} \frac{\sin 3x} {x}.\end{align*}

Solution:

Example 5:

Evaluate \begin{align*} \lim_{x \rightarrow \pi/2} \frac{5 - 5 \sin x} {\cos x}\end{align*} .

Solution:

We can use L’Hospital’s rule since the limit produces the \begin{align*}0/0\end{align*} once \begin{align*}x = \pi/2\end{align*} is substituted. Hence

A broader application of L’Hospital’s rule is when \begin{align*}x = a\end{align*} 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 \begin{align*} \lim_{x \rightarrow 0} \frac{e^x -\cos x} {x^2}.\end{align*}

Solution:

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:

## Review Questions

Find the limits.

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

1. \begin{align*}1\end{align*}
2. \begin{align*}1/\pi\end{align*}
3. \begin{align*}4\end{align*}
4. \begin{align*}-1\end{align*}
5. \begin{align*}-1\end{align*}
6. \begin{align*}\ln x\end{align*}
7. \begin{align*} \frac{\pi} {4}\end{align*}

## 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

Aug 19, 2015