Evaluate the following limits:

$=\lim _{x \rightarrow 0} \frac{e^{\tan x}-1}{x}$

$=\lim _{x \rightarrow 0} \frac{e^{\tan x}-1}{x} \times \frac{\tan x}{\tan x}$

$=\lim _{x \rightarrow 0} \frac{e^{\tan x}-1}{\tan x} \times \frac{\tan x}{x}$

$=1 \times 1$

$=1$

$\therefore \lim _{x \rightarrow 0} \frac{e^{\tan x}-1}{x}=1$