What is $\lim_{x \to 0} e^{\frac{\tan 2x}{x}}$?

Explanation

1. 01

   $e^{u}$ is continuous, so the limit passes inside: the answer is $e^{L}$ with $L = \lim_{x \to 0} \frac{\tan 2x}{x}$.

2. 02

   $\frac{\tan 2x}{x} = 2 \cdot \frac{\tan 2x}{2x} \to 2$, so the limit is $e^{2}$.