<meta http-equiv="refresh" content="1; url=/nojavascript/"> L’Hospital’s Rule | CK-12 Foundation
Dismiss
Skip Navigation
You are reading an older version of this FlexBook® textbook: CK-12 Calculus Go to the latest version.

6.6: L’Hospital’s Rule

Created by: CK-12
 0  0  0

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 ].

Review Answers

  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.

Image Attributions

Description

Categories:

Grades:

Date Created:

Feb 23, 2012

Last Modified:

Aug 21, 2014
Files can only be attached to the latest version of None

Reviews

Please wait...
Please wait...
Image Detail
Sizes: Medium | Original
 
CK.MAT.ENG.SE.1.Calculus.6.6
ShareThis Copy and Paste

Original text