Find $\lim_\limits{x\to 0} \frac{\sin x}{e^x -1 -\sin x}$ I have to find $$\lim_\limits{x\to 0} \dfrac{\sin x}{e^x -1 -\sin x}.$$
By using L'Hopital, $$\lim_\limits{x\to 0} \frac{\sin x}{e^x -1 -\sin x}= \lim_\limits{x\to 0} \frac{\cos x}{e^x -\cos x}.$$
Wolfram now states that $\nexists$ because $\lim_\limits{x\to 0^+} \frac{\sin x}{e^x -1 -\sin x}=\infty$ and $\lim_\limits{x\to 0^-} \frac{\sin x}{e^x -1 -\sin x}=-\infty$,
but I don't see why $\lim_\limits{x\to 0^-} \frac{\sin x}{e^x -1 -\sin x}=-\infty$

Use equivalents and Taylor's formula at order $2$:


*

*$\sin x\sim_0x$,

*$\mathrm e^x-1-\sin x=1+ x+\dfrac{x^2}2+o(x^2)-1-x-o(x^2)=\dfrac{x^2}2+o(x^2)$, so $\mathrm e^x-1-\sin x\simeq_0\dfrac{x^2}2.$


Thus we have
$$\frac{\sin x}{e^x -1 -\sin x}\simeq_0\frac x{\dfrac{x^2}2}=\frac 2x\to\begin{cases}+\infty&\text{if }x\to 0^+\\ -\infty&\text{if }x\to 0^-
\end{cases}$$
