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: $$
\underbrace{f'(0) = \lim_{h\to0} \frac{f(0+h)-f(0)} h}_{\text{definition of ``derivative"}} = \lim_{h\to0}\frac{e^{2\sin h} - e^{2\sin 0}} h = \lim_{h\to0}\frac{e^{2\sin h} - 1} h.
$$
So just find $f'(0)$ by the methods you would normally use to compute a derivative.
A: 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.
A: Hint: Consider the function $f(x)=e^{2\sin x}$. What is the derivative of said function at $x=0$?
A: HINT: $\lim_{y\to0}\frac{e^y-1}{y}=1$. Hence
$$
\lim_{x\to0}\frac{e^{2\sin x}-1}{x}=
\lim_{x\to0}\frac{e^{2\sin x}-1}{2\sin x}\cdot\frac{2\sin x}{x}
$$
and the rest should be easy.
A: When given a task that includes the text "You are not allowed to use so-and-so Rule", you can still use the proof of that rule in your answer.
So, look up the proof of L'H in your text book or on Wikipedia, put in your function where apropriate. Ignore everything that isn't relevant to your case.
In the end you will have your answer as well as a better understanding of the rule.
A: I guess the answer should be $2$. 
Multiply up and down by $2 \sin x$. Since we know $\lim_{x \to0} \frac{e^x -1}{x} =1$. Proceeding that way leaves us only with $\lim_{x \to0} \frac{2 \sin x}{x}$ (and we know $\lim_{x \to 0} \frac{\sin x}{x}=1$) So answer should be $2$.
A: With $x$ in the neigborhood of $0$: $$e^{2\sin(x)}=1+2x+o(x^2)$$ so $$\frac{e^{2\sin(x)}-1}{x}=2+o(x)$$.
A: 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  
$$\lim_{x\to0}\frac{1+x-1}{x}=1$$
