L'Hopital's Rule and The Indeterminate Forms of Limits in Calculus

L'Hopital'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.

L'Hopital's Rule

If \( \lim_{{x \to a}} f(x) = 0 \) and \( \lim_{{x \to a}} g(x) = 0 \) 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

Example 1

Find the limit \( \lim_{{x \to 0}} \dfrac{{\sin x}}{{x}} \)

Solution to Example 1:

Since \[ \lim_{{x \to 0}} \sin x = 0 \] and \[ \lim_{{x \to 0}} x = 0 \] L'Hopital'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 \( \lim_{{x \to 0}} \dfrac{{e^x - 1}}{{x}} \)

Solution to Example 2:

Note that \[ \lim_{{x \to 0}} (e^x - 1) = 0 \] and \[ \lim_{{x \to 0}} x = 0 \] We can use L'Hopital'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 \( \lim_{{x \to 1}} \dfrac{{x^2 - 1}}{{x - 1}} \)

Solution to Example 3:

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'Hopital'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 \( \lim_{{x \to 2}} \dfrac{{\ln(x - 1)}}{{x - 2}} \)

Solution to Example 4:

Limit of numerator \[ \lim_{{x \to 2}} \ln(x - 1) = 0 \] Limit of denominator \[ \lim_{{x \to 2}} (x - 1) = 0 \] Both limits are equal to zero, L'Hopital's rule may be used \[ \lim_{{x \to 2}} \dfrac{{\ln(x - 1)}}{{x - 2}} = \lim_{{x \to 2}} \dfrac{\dfrac{{d(\ln(x - 1))}}{{dx}}} { \dfrac{{d(x - 2)}}{{dx}} } \] \[ = \lim_{{x \to 2}} \dfrac{{1/(x-1)}}{{1}} = \dfrac{{1/(2 -1)}}{{1}} = 1 \]

Example 5

Find the limit \( \lim_{{x \to 0}} \dfrac{{1 - \cos x}}{{6x^2}} \)

Solution to Example 5:

Limit of numerator and denominator \[ \lim_{{x \to 0}} (1 - \cos x) = 0 \] \([ \lim_{{x \to 0}} 6x^2 = 0 \] L'Hopital'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'Hopital's rulea second time \[ = \lim_{{x \to 0}} \dfrac{{\cos x}}{{12}} = \dfrac{{\cos 0}}{{12}} = \dfrac{1}{12} \]

Exercises

Find the limits
1. \( \quad \lim_{{x \to 0}} \dfrac{{\sin 4x}}{{\sin 2x}} \)
2. \( \quad \lim_{{x \to 0}} \dfrac{{\tan x}}{{x}} \)
3. \( \quad \lim_{{x \to 1}} \dfrac{{\ln x}}{{3x - 3}} \)
4. \( \quad \lim_{{x \to 0}} \dfrac{{e^x - 1}}{{\sin 2x}} \)

Solutions to Above Exercises

1. \( \quad 2 \)
2. \( \quad 1 \)
3. \( \quad \dfrac{1}{3} \)
4. \( \quad \dfrac{1}{2} \)