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.
[Math] How to find $\lim_{x \to 0}\left(\frac{e^{2\sin x}-1}{x}\right)$ without l’Hopital’s Rule
calculuslimitslimits-without-lhopital
Best Answer
Hint: Consider the function $f(x)=e^{2\sin x}$. What is the derivative of said function at $x=0$?