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

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

\begin{align*}\lim_{x \to \ a} \frac {f(x)}{g(x)}\end{align*}limx af(x)g(x)

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

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

Example 1:

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

Solution:

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

Therefore L’Hospital’s rule applies:

\begin{align*}\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}.\end{align*}

Example 2:

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

Solution:

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

Using L’Hospital’s rule,

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

Example 3:

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

Solution:

\begin{align*}\lim_{x \rightarrow 3} = \frac{x^2 - 9} {x - 3} = \lim_{x \rightarrow 3} \frac{2x} {1} = 6.\end{align*}

Example 4:

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

Solution:

\begin{align*}\lim_{x \rightarrow 0} \frac{\sin 3x} {x} = \lim_{x \rightarrow 0} \frac{3 \cos 3x} {1} = 3.\end{align*}

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

\begin{align*}\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.\end{align*}

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:

\begin{align*}\lim_{x \rightarrow 0} \frac{e^x - \cos x} {x^2} = \lim_{x \rightarrow 0} \frac{e^x + \sin x} {2x}.\end{align*}

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:

\begin{align*}& = \lim_{x \rightarrow 0} \frac{e^x - \cos x} {2}\\ & = \frac{1 - 1} {2} = \frac{0} {2} = 0.\end{align*}

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 \begin{align*} \frac{f'(c)} {g'(c)} = \frac{f(b) - f(a)} {g(b) - g(a)}.\end{align*} Find all possible values of \begin{align*}c\end{align*} in the interval \begin{align*}(a, b)\end{align*} that satisfy this property for \begin{align*}f(x) & = \cos x\\ g(x) & = \sin x\end{align*} on the interval \begin{align*} [a, b] = \left [0, \frac{\pi} {2}\right ].\end{align*}

Review Answers

  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.

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Feb 23, 2012
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CK.MAT.ENG.SE.1.Calculus.6.6