$\lim_{x \to 0} \frac{e^{\tan x} - e^x}{\tan x - x}$ is equal to
Step-by-Step Solution
Key Concept: When direct substitution gives an indeterminate form, L'Hôpital's rule can be applied repeatedly until the limit becomes determinate.
We evaluate $L = \lim_{x \to 0} \frac{\sin x - \cos x}{x^2}$, which is $\frac{0}{0}$ form. Applying L'Hôpital's rule: $\lim_{x \to 0} \frac{\cos x + \sin x}{2x}$, still $\frac{0}{0}$ form. Apply L'Hôpital's rule again: $\lim_{x \to 0} \frac{-\sin x + \cos x}{2} = \frac{0 + 1}{2} = \frac{1}{2}$.
Correct Answer: 1/2