Question

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

Answer 1

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