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.
Hint: Consider the function $f(x)=e^{2\sin x}$. What is the derivative of said function at $x=0$?
Well, there is an easy solution if you know the series expansion of $e^x$.

Near $x=0$,  $\sin(2x) \approx 2x$ therefor the given equation reduces to 

$$\lim_{x\to0} \frac{e^{2x}-1}{x}$$

Now apply series expansion for $e^{2x}=1+x+...$ , neglecting higher powers as $x\to0$
There  for it reduces to  

$$\lim_{x\to0}\frac{1+x-1}{x}=1$$