[tex]\displaystyle\lim_{x\to\pi}\dfrac{e^{\sin x}-1}{x-\pi}[/tex]
Notice that if [tex]f(x)=e^{\sin x}[/tex], then [tex]f(\pi)=e^{\sin\pi}=e^0=1[/tex]. Recall the definition of the derivative of a function [tex]f(x)[/tex] at a point [tex]x=c[/tex]:
[tex]f'(c):=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}[/tex]
So the value of this limit is exactly the value of the derivative of [tex]f(x)=e^{\sin x}[/tex] at [tex]x=\pi[/tex].
You have
[tex]f'(x)=\cos x\,e^{\sin x}\implies f'(\pi)=\cos\pi\,e^{\sin\pi}=-1[/tex]
