Editorial StaffL’Hôpital’s Rule provides a method for evaluating limits involving indeterminate forms.
By taking the ratio of the derivatives of the functions involved, we can often simplify the evaluation of such limits.
This rule is especially useful when direct substitution or other algebraic methods fail to determine the limit.
It is used to evaluate limits involving indeterminate forms, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$.
It states that under certain conditions, the limit of the ratio of two functions can be found by taking the ratio of their derivatives.
Statement of L’Hôpital’s Rule
Let $f(x)$ and $g(x)$ be differentiable functions defined on an open interval containing $a$, except possibly at $a$ itself. If
\[
\lim_{x \to a} f(x) = 0 \quad \text{and} \quad \lim_{x \to a} g(x) = 0
\]
or
\[
\lim_{x \to a} f(x) = \pm \infty \quad \text{and} \quad \lim_{x \to a} g(x) = \pm \infty,
\]
and
\[
\lim_{x \to a} \frac{f'(x)}{g'(x)} \quad \text{exists or is} \quad \pm \infty,
\]
then
\[
\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}.
\]
Example
Let’s consider the limit
\[
L = \lim_{x \to 0} \frac{\sin x}{x}.
\]
This limit is of the form $\frac{0}{0}$ as $x$ approaches 0. We can apply L’Hôpital’s Rule to evaluate this limit.
\begin{align*}
\lim_{x \to 0} \frac{\sin x}{x} &= \lim_{x \to 0} \frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(x)} \\
&= \lim_{x \to 0} \frac{\cos x}{1} \\
&= \cos 0 \\
&= 1.
\end{align*}
Therefore, the limit $\lim_{x \to 0} \frac{\sin x}{x}$ is equal to 1.
