Question:
Evaluate the following limits:
$\lim _{x \rightarrow 0} \frac{\left(e^{\tan x}-1\right)}{\tan x}$
Solution:
$=\lim _{x \rightarrow 0} \frac{\left(e^{\tan x}-1\right)}{\tan x}$
As x tends to 0, tan(x) also tends to zero,
So,
$\lim _{x \rightarrow 0} \frac{\left(e^{\tan x}-1\right)}{\tan x}=\lim _{\tan x \rightarrow 0} \frac{\left(e^{\tan x}-1\right)}{\tan x}$
$=1$
$\therefore \lim _{x \rightarrow 0} \frac{\left(e^{\tan x}-1\right)}{\tan x}=1$
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