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?
 A: Your calculations are correct except for your final result which should obviously be only $+\infty$.
Here another way which uses the standard limits $\lim_{t\to 0}\frac{e^t-1}{t}= 1$ and $\lim_{t\to 0}\frac{\sin t}{t}=1$.
First of all note that for $x>0$ you have $\sin x < x$ and for $x<0$ you have $x < \sin x$. Hence
$$\frac{\sin x}{e^x - e^{\sin x}} >0 \text{ for } x \in \left[-\frac{\pi}2 , \frac{\pi}2\right]\setminus\{0\}$$
Now, just consider the reciprocal
\begin{eqnarray*} \frac{e^x - e^{\sin x}}{\sin x}
& = &  \frac{e^x -1 - \left(e^{\sin x}-1\right)}{\sin x} \\
& = & \frac{e^x -1}{x}\cdot \frac{x}{\sin x} - \frac{e^{\sin x}-1}{\sin x} \\
& \stackrel{x \to 0}{\longrightarrow} & 1\cdot 1 - 1 = 0
\end{eqnarray*}
Hence,
$$\lim_{x\to 0} \frac{\sin x}{e^x - e^{\sin x}} = +\infty$$
A: 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.
A: We have
\begin{align}
&\lim_{x\to 0} \frac{\sin(x)}{\exp(x)-\exp(\sin(x))}
=\lim_{x\to 0}\frac{\sin(x)}{x}\cdot\frac{1}{\exp(\sin(x))}\cdot \frac{x}{\exp(x-\sin(x))-1}\\
&=\lim_{x\to 0}\frac{x}{\exp(x-\sin(x))-1}\stackrel{\text{L'Hopital}}=\lim_{x\to 0}\frac{1}{\exp(x-\sin(x))(1-\cos(x))}\\
&=\lim_{x\to 0} \frac{1}{1-\cos(x)} = +\infty
\end{align}
A: As stated in the comments, your solution is correct up until the last step where the limit should be $+\infty$.
Since $f(x)=x−\sin x$ is increasing in a neighborhood of $0$ we can invert this function and use $\lim _{ x\rightarrow 0 }\frac { e^x -1 }{ x } =1 $ to form
$$\lim _{ x\to 0 } \frac { x-\sin  x }{ e^x -e^{ \sin  x } } =\lim _{ x\to 0 } \frac {1}{ \dfrac{{ e^{ \sin x } \left( e^{ x-\sin x }-1 \right) }}{x-\sin  x} } =\lim _{ x\rightarrow 0 }{ \frac{1}{e^{ \sin x }} } =1 .$$
Then since $\lim _{ x\rightarrow 0 }\frac { x}{ \sin x } =1 $
$$\lim _{ x\to 0 }\frac { \sin  x }{x-\sin x}=\lim _{ x\to 0 }\frac{1}{\frac{x}{\sin x}-1}=+\infty.$$
Therefore
$$\lim _{ x\to 0 } \frac { \sin  x }{ e^x -e^{ \sin  x } }=\left(\lim _{ x\to 0 }\frac { \sin  x }{x-\sin x}\right)\left(\lim _{ x\to 0 }\frac { x-\sin  x }{ e^x -e^{ \sin  x } }\right)=+\infty.$$
