L'Hôpital’s Theorem
Let \( f \) and \( g \) be two functions that are differentiable in a neighborhood of \( x_0 \), such that: $$ \lim_{x \rightarrow x_0} f(x) = 0 $$ $$ \lim_{x \rightarrow x_0} g(x) = 0 $$ or: $$ \lim_{x \rightarrow x_0} f(x) = \infty $$ $$ \lim_{x \rightarrow x_0} g(x) = \infty $$ and suppose that \( g(x) \neq 0 \) and \( g'(x) \neq 0 \) in that neighborhood: $$ g(x) \neq 0 \quad \text{and} \quad g'(x) \neq 0 $$ If the following limit exists and is finite: $$ \lim_{x \rightarrow x_0} \frac{f'(x)}{g'(x)} = L \in \mathbb{R} $$ then it follows that: $$ \lim_{x \rightarrow x_0} \frac{f(x)}{g(x)} = L \in \mathbb{R} $$
It is not necessary for the functions to be differentiable *at* \( x_0 \); it is sufficient for them to be differentiable in a neighborhood of \( x_0 \).
Here, \( x_0 \) may be either a finite point or infinity.
Note: if applying L’Hôpital’s Rule once still results in an indeterminate form of type \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), the rule can be applied repeatedly, using higher-order derivatives, until the limit can be evaluated.
What is L'Hôpital’s Theorem used for?
It provides a powerful method for evaluating limits of indeterminate form \( 0/0 \):
$$ \lim_{x \rightarrow x_0} \frac{f(x)}{g(x)} = \frac{0}{0} $$
It can also be used for limits of type \( \infty/\infty \):
$$ \lim_{x \rightarrow x_0} \frac{f(x)}{g(x)} = \frac{\infty}{\infty} $$
In this case, both \( f(x) \) and \( g(x) \) tend to infinity.
Note. Other indeterminate forms - such as \( \infty - \infty \), \( 0 \cdot \infty \), \( 0^0 \), \( 1^\infty \), \( \infty^0 \) - must first be rewritten as \( 0/0 \) or \( \infty/\infty \) to apply L’Hôpital’s Theorem. For instance, products can be written as quotients: $$ f \cdot g = \frac{f}{\frac{1}{g}} $$ or differences can be written as quotients: $$ f - g = \frac{ \frac{1}{g} - \frac{1}{f} }{ \frac{1}{f \cdot g} } $$ provided the resulting form is of type \( 0/0 \) or \( \infty/\infty \).
Examples and Exercises
Example 1
The following limit is an indeterminate form of type \( 0/0 \):
$$\lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 0} \frac{e^{2x} - e^{-2x}}{ \sin 5x } = \frac{0}{0} $$
Since both limits tend to zero, L’Hôpital’s conditions are satisfied:
$$ \lim_{x \rightarrow 0} f(x) = 0 \\ \lim_{x \rightarrow 0} g(x) = 0 $$
We differentiate numerator and denominator:
$$ f'(x) = 2e^{2x} + 2e^{-2x} $$ $$ g'(x) = 5 \cos 5x $$
Then evaluate the limit of their ratio:
$$ \lim_{x \rightarrow 0} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow 0} \frac{2e^{2x} + 2e^{-2x}}{5 \cos 5x} = \frac{4}{5} $$
This is the value of the original limit.
What if the result remains indeterminate? If the first derivative still yields an indeterminate form \( 0/0 \) or \( \infty/\infty \), we can apply L’Hôpital’s Rule again using higher derivatives. The theorem applies to derivatives of any order.
Example 2
Here is an indeterminate form \( \infty/\infty \):
$$\lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \infty} \frac{e^x}{x} = \frac{\infty}{\infty} $$
Applying L’Hôpital’s Rule:
$$ \lim_{x \rightarrow \infty} \frac{f'(x)}{g'(x)} = \lim_{x \rightarrow \infty} \frac{e^x}{1} = \infty $$
Thus, the original limit also tends to infinity:
$$ \lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} = \infty $$
Example 3
This is an indeterminate form \( 0 \cdot \infty \):
$$\lim_{x \rightarrow 0^+} x^2 \cdot \log x = 0 \cdot (-\infty) $$
Rewriting as a quotient:
$$\lim_{x \rightarrow 0^+} \frac{ \log x }{ \frac{1}{x^2} } $$
$$\lim_{x \rightarrow 0^+} \frac{ \log x }{ x^{-2} } $$
This is now \( \infty/\infty \), so we apply L’Hôpital’s Rule:
$$\lim_{x \rightarrow 0^+} \frac{ \frac{1}{x} }{ -2x^{-3} } $$
$$\lim_{x \rightarrow 0^+} -2x^2 = 0 $$
Example 4
Another example of \( \infty/\infty \):
$$\lim_{x \rightarrow \infty} \frac{x^2}{e^x} = \frac{\infty}{\infty} $$
Applying L’Hôpital’s Rule once more:
$$\lim_{x \rightarrow \infty} \frac{2x}{e^x} = \frac{\infty}{\infty} $$
Since L’Hôpital’s Rule applies to higher derivatives as well, we proceed:
$$\lim_{x \rightarrow \infty} \frac{2}{e^x} = 0 $$
Thus, the limit converges to zero.
The Proof
Let \( f(x) \) and \( g(x) \) be differentiable in a neighborhood of \( x_0 \), with \( g(x) \neq 0 \), and: $$ \lim_{x \rightarrow x_0} f(x) = 0 $$ $$ \lim_{x \rightarrow x_0} g(x) = 0 $$
Since \( f \) and \( g \) are differentiable, they are also continuous in a neighborhood of \( x_0 \), so: $$ f(x_0) = g(x_0) = 0 $$
We rewrite the limit as:
$$ \lim_{x \rightarrow x_0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{g(x) - g(x_0)} $$
Dividing numerator and denominator by \( x - x_0 \):
$$ \lim_{x \rightarrow x_0} \frac{ \frac{f(x) - f(x_0)}{x - x_0} }{ \frac{g(x) - g(x_0)}{x - x_0} } $$
The limit becomes:
$$ \frac{ f'(x_0) }{ g'(x_0) } = \lim_{x \rightarrow x_0} \frac{f'(x)}{g'(x)} $$
Thus, we establish: $$ \lim_{x \rightarrow x_0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow x_0} \frac{f'(x)}{g'(x)} $$
And so on.
