Determine $\lim\limits_{x\to 0, x\neq 0}\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{1}{2}$ without using L'Hospital How to prove that
$$\lim\limits_{x\to 0, x\neq 0}\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{1}{2}$$ 
without using L'Hospital?
Using L'Hospital, it's quite easy. But without, I don't get this. I tried different approaches, for example writing $$e^{\sin(x)}=\sum\limits_{k=0}^\infty\frac{\sin(x)^k}{k!}$$
and
$$\sin(2x)=2\sin(x)\cos(x)$$
and get
$$\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{\sin(x)+\sum\limits_{k=2}^\infty\frac{\sin(x)^k}{k!} }{2\sin(x)\cos(x)}$$
but it seems to be unrewarding. How can I calculate the limit instead?
Any advice will be appreciated.

Hint: If I continue with your approach
$$\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{\sin(x)+\sum\limits_{k=2}^\infty\frac{\sin(x)^k}{k!} }{2\sin(x)\cos(x)}=\frac{1+\sum\limits_{k=2}^\infty\frac{\sin(x)^{k-1}}{k!} }{2\cos(x)}$$
You can now set $x=0$, but you'll need to justify this step.
