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.

$$
\underbrace{f'(0) = \lim_{h\to0} \frac{f(0+h)-f(0)} h}_{\text{definition of ``derivative"}} = \lim_{h\to0}\frac{e^{2\sin h} - e^{2\sin 0}} h = \lim_{h\to0}\frac{e^{2\sin h} - 1} h.
$$
So just find $f'(0)$ by the methods you would normally use to compute a derivative.
