L'Hôpital's Rule

If $\lim_{x \to a} \frac{f(x)}{g(x)}$ gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$:

$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$

(if the right side exists).

$\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1$.

$\lim_{x \to \infty} \frac{x^2}{e^x} = \lim \frac{2x}{e^x} = \lim \frac{2}{e^x} = 0$.

$\lim_{x \to 0} \frac{e^x - 1}{x} = \lim \frac{e^x}{1} = 1$.

• Form is not indeterminate. $\frac{5}{0}$ is not $\frac{0}{0}$ — it's just $\pm\infty$.
• Apply algebraic simplification first when possible.