Answer
$\begin{aligned}
&\lim\limits_{x\to\pi}\frac{e^{\sin x}-1}{x-\pi}\\
=&\lim\limits_{x\to\pi}\frac{\cos xe^{\sin x}}{1} \\
=&-e^{\sin\pi} \\
=&-1
\end{aligned}$
Work Step by Step
$\begin{aligned}
&\lim\limits_{x\to\pi}\frac{e^{\sin x}-1}{x-\pi}\\
=&\lim\limits_{x\to\pi}\frac{\cos xe^{\sin x}}{1} \\
=&-e^{\sin\pi} \\
=&-1
\end{aligned}$
