Question

$\underset{x\to 0}{\lim }\frac{e^x-1+\sin x}{x}$

Answer 1

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{e^x-1+\sin x}{x}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{e^x-1}{x}+\frac{\sin x}{x}\right] \\ =1+1 \\ =2 \end{gathered}$$