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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