[Math] $\lim_{x \to \pi}\frac{e^{\sin x} – 1}{x – \pi}$

Question

Please don't give me the answer – I only want a hint.
$$\lim_{x \to \pi}\frac{e^{\sin x} – 1}{x – \pi}$$
This is a "Problems Plus" question from Stewart's Early Transcendentals (specifically, chapter 3 question 15).

I have no idea how to solve this limit – I didn't even think it existed at first because the limit of the numerator is $-1$ and the limit of the denominator is $0$, but Wolfram Alpha goes against this logic in saying that the limit is $-1$.

I've tried transforming this limit into the definition of a derivative for an easier time but have failed.

Any help is appreciated.

EDIT: This was in a section before L'Hoptial's Rule. However, I just realized the answer to the question – my apologies, I kind of forgot $e^0 = 1$ and not $0$.

Answer 1

$$\lim_{x\to\pi}\dfrac{e^{\sin(x)}-e^{\sin \pi}}{x-\pi}$$ Do you recognize the derivative's limit definition here? (The answer is $\cos \pi\times e^{\sin \pi}=-1$.)