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?
Best Answer
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,
$$\lim_{x\to 0} \frac{\sin x}{e^x - e^{\sin x}} = +\infty$$