How can I find $\lim_{x \to 0}\left(\frac{e^{2\sin x}-1}{x}\right)$ without l'Hopital's Rule? How do I evaluate
$$\lim_{x \to 0}\left(\frac{e^{2\sin x}-1}{x}\right)$$
I know it's the indeterminate form since the numerator and denominator both approach 0, but I can't use l'Hopital's rule so I'm not sure how to go about finding the limit.

A definition of the derivative is at $x=a$ is $$f'(a)=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$ Now plug in $a=0,f(x)=e^{2\sin x}$ and you'll have your answer.
