L'Hôpital's rule allows us to replace a limit problem with another that may be simpler to solve. Several examples are presented along with their solutions and detailed explanations.
In what follows, \( C \) represents the constant of integration where applicable.
L'Hôpital's Rule
If \( \lim_{x \to a} f(x) = 0 \) and \( \lim_{x \to a} g(x) = 0 \) (or both approach \( \pm\infty \)) and if \( \lim_{x \to a} \dfrac{f'(x)}{g'(x)} \) has a finite value \( L \), or is of the form \( +\infty \) or \( -\infty \), then:
\[ \lim \dfrac{f(x)}{g(x)} = \lim \dfrac{f'(x)}{g'(x)} \]where \( \lim \) stands for \( \lim_{x \to a} \), \( \lim_{x \to a^+} \), \( \lim_{x \to a^-} \), \( \lim_{x \to +\infty} \), or \( \lim_{x \to -\infty} \).
\( f'(x) \) and \( g'(x) \) are the derivatives of \( f(x) \) and \( g(x) \) respectively.
Examples with Detailed Solutions
Click on each example to view its detailed step-by-step solution.
Example 1
Find the limit \( \displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} \)
Show Solution
Since:
\[ \lim_{x \to 0} \sin x = 0 \]and
\[ \lim_{x \to 0} x = 0 \]L'Hôpital's rule can be used to evaluate the above limit as follows:
\[ \lim_{x \to 0} \dfrac{\sin x}{x} = \lim_{x \to 0} \dfrac{\dfrac{d(\sin x)}{dx}}{\dfrac{d(x)}{dx}} \] \[ = \lim_{x \to 0} \dfrac{\cos x}{1} = \dfrac{\cos 0}{1} = 1 \]Example 2
Find the limit \( \displaystyle \lim_{x \to 0} \dfrac{e^x - 1}{x} \)
Show Solution
Note that:
\[ \lim_{x \to 0} (e^x - 1) = 0 \]and
\[ \lim_{x \to 0} x = 0 \]We can use L'Hôpital's rule to calculate the given limit as follows:
\[ \lim_{x \to 0} \dfrac{e^x - 1}{x} = \lim_{x \to 0} \dfrac{\dfrac{d(e^x - 1)}{dx}}{\dfrac{d(x)}{dx}} \] \[ = \lim_{x \to 0} \dfrac{e^x}{1} = \dfrac{e^0}{1} = 1 \]Example 3
Find the limit \( \displaystyle \lim_{x \to 1} \dfrac{x^2 - 1}{x - 1} \)
Show Solution
Since the limit of the numerator:
\[ \lim_{x \to 1} (x^2 - 1) = 0 \]and that of the denominator:
\[ \lim_{x \to 1} (x - 1) = 0 \]are both equal to zero, we can use L'Hôpital's rule to calculate the limit:
\[ \lim_{x \to 1} \dfrac{x^2 - 1}{x - 1} = \lim_{x \to 1} \dfrac{\dfrac{d(x^2 - 1)}{dx}}{\dfrac{d(x - 1)}{dx}} \] \[ = \lim_{x \to 1} \dfrac{2x}{1} = \dfrac{2(1)}{1} = 2 \]Note that the same limit may be calculated by first factoring as follows:
\[ \lim_{x \to 1} \dfrac{x^2 - 1}{x - 1} = \lim_{x \to 1} \dfrac{(x - 1)(x + 1)}{x - 1} = \lim_{x \to 1} (x + 1) = 2 \]Example 4
Find the limit \( \displaystyle \lim_{x \to 2} \dfrac{\ln(x - 1)}{x - 2} \)
Show Solution
Limit of numerator:
\[ \lim_{x \to 2} \ln(x - 1) = \ln(1) = 0 \]Limit of denominator:
\[ \lim_{x \to 2} (x - 2) = 0 \]Both limits are equal to zero, so L'Hôpital's rule may be used:
\[ \lim_{x \to 2} \dfrac{\ln(x - 1)}{x - 2} = \lim_{x \to 2} \dfrac{\dfrac{d(\ln(x - 1))}{dx}}{\dfrac{d(x - 2)}{dx}} \] \[ = \lim_{x \to 2} \dfrac{\dfrac{1}{x - 1}}{1} = \dfrac{\dfrac{1}{2 - 1}}{1} = 1 \]Example 5
Find the limit \( \displaystyle \lim_{x \to 0} \dfrac{1 - \cos x}{6x^2} \)
Show Solution
Limit of numerator and denominator:
\[ \lim_{x \to 0} (1 - \cos x) = 0 \quad \text{and} \quad \lim_{x \to 0} 6x^2 = 0 \]L'Hôpital's rule may be used:
\[ \lim_{x \to 0} \dfrac{1 - \cos x}{6x^2} = \lim_{x \to 0} \dfrac{\sin x}{12x} \]The new limit is also indeterminate \( \dfrac{0}{0} \), and we may apply L'Hôpital's rule a second time:
\[ = \lim_{x \to 0} \dfrac{\cos x}{12} = \dfrac{\cos 0}{12} = \dfrac{1}{12} \]Exercises
Calculate the following limits. Click each exercise to check your answers.
Exercise 1
\( \displaystyle \lim_{x \to 0} \dfrac{\sin 4x}{\sin 2x} \)
Show Answer
Exercise 2
\( \displaystyle \lim_{x \to 0} \dfrac{\tan x}{x} \)
Show Answer
Exercise 3
\( \displaystyle \lim_{x \to 1} \dfrac{\ln x}{3x - 3} \)
Show Answer
Exercise 4
\( \displaystyle \lim_{x \to 0} \dfrac{e^x - 1}{\sin 2x} \)