Evaluate the following limits:

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

As x tends to 0, tan(x) also tends to zero,

So,

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

$=1$

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