calculus - Determine $limlimits_{xto 0, xneq 0}frac{e^{sin(x)}-1}{sin(2x)}=frac{1}{2}$ 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.

Answer

From the known limit
$$
\lim\limits_{u\to 0}\frac{e^u-1}{u}=1,
$$ one gets
$$
\lim\limits_{x\to 0}\frac{e^{\sin x}-1}{\sin(2x)}=\lim\limits_{x\to 0}\left(\frac{e^{\sin x}-1}{\sin x}\cdot\frac{\sin x}{\sin(2x)}\right)=\lim\limits_{x\to 0}\left(\frac{e^{\sin x}-1}{\sin x}\cdot\frac{1}{2\cos x}\right)=\color{red}{1}\cdot\frac{1}{2}.

$$