Evaluating $\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}$? 
Calculate this limit
$$\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}$$

My attempt: using the limit development : we find
$$\exp(\sin(x)-x)=\exp(x-\frac{x^{3}}{3!}+o (x^{3})-x)=\exp(-\frac{x^3}{3!}+o(x^3))=1-\frac{x^3}{3!}+o(x^3)$$
So:
\begin{align}
\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}
&=\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)}\left(\frac{1}{\frac{x^3}{3!}+o(x^3)}\right)\\\
& \sim \lim\limits_{x\rightarrow 0}\left(\frac{\sin(x)}{x}\right)\left(\frac{3!}{x^{2}\exp(x)}\right)=\pm\infty.
\end{align}
Is this correct?

You don't really need to factor like that. You can just say $\sin(x)=x-x^3/6+o(x^3)$ so
$$\exp(\sin(x))=1+\sin(x)+\sin(x)^2/2+\sin(x)^3/6+o(\sin(x)^3) \\
= 1 + (x-x^3/6) + (x-x^3/6)^2/2 + (x-x^3/6)^3/6 + o(x^3) \\
= 1 + x + x^2/2 - x^3/6 + x^3/6 + o(x^3) \\
= 1 + x + x^2/2 + o(x^3)$$
so $\exp(x)-\exp(\sin(x))=x^3/6 + o(x^3)$.
Your method is a little bit easier in this particular problem but it is pretty limited in its scope compared to this method.
The one actual issue in your calculation is that $\sin(x)/x \to 1$ from either side and $1/x^2 \to +\infty$ from either side, so you have $+\infty$ on both sides.
