Question

Limit question: lim x-->pi ((e^sinx)-1)/(x-pi)

Answer 1

$\displaystyle\lim_{x\to\pi}\dfrac{e^{\sin x}-1}{x-\pi}$

Notice that if $f(x)=e^{\sin x}$, then $f(\pi)=e^{\sin\pi}=e^0=1$. Recall the definition of the derivative of a function $f(x)$ at a point $x=c$:

$f'(c):=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

So the value of this limit is exactly the value of the derivative of $f(x)=e^{\sin x}$ at $x=\pi$.

You have

$f'(x)=\cos x\,e^{\sin x}\implies f'(\pi)=\cos\pi\,e^{\sin\pi}=-1$