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 $\lim_\limits{x\to 0^-} \frac{\sin x}{e^x -1 -\sin x}=-\infty$

Hint:
\begin{align}
& \lim_{x\to 0} \frac{\sin x}{e^x -1 -\sin x} \\[6pt]
= {} & \lim_{x\to 0} \frac{\sin x}{1+x+\frac{x^2}{2}+o(x^3) -1 -(x-\frac{x^3}{6}+o(x^5))} \\[6pt]
= {} & \lim_{x\to 0} \frac{\sin x}{\frac{x^2}{2}+o(x^3) } \\[6pt]
= {} & \lim_{x\to 0} \frac{x}{\frac{x^2}{2}+o(x^3) } \\[6pt]
= {} & \lim_{x\to 0} \frac{1}{\frac{x}{2}+o(x^2) }
\end{align}
