Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1+\sin\text{x}}{\text{x}}$

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1+\sin\text{x}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}+\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}$
$=\text{log e+1}$
$=2$

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