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 $\lim_\limits{x\to 0^-} \frac{\sin x}{e^x -1 -\sin x}=-\infty$
 A: Intuitively, you can use the approximations $\sin x\approx x$ and $e^x\approx1+x+x^2/2$, so that the given expression behaves like $2/x$, which has no limit.
A: $$\lim_{x\to 0} \frac{\sin x}{e^x -1 -\sin x}= \lim_{x\to 0} \frac{\cos x}{e^x  -\cos x}=\infty$$
because $e^x-\cos{x}\rightarrow0$ and $\cos{x}\rightarrow1$.
$$ \lim_{x\to 0^+} \frac{\cos x}{e^x  -\cos x}=+\infty$$ because $e^x  -\cos x=e^x-1+1-\cos{x}>0$ for $x\rightarrow0^+$.
$$ \lim_{x\to 0^-} \frac{\cos x}{e^x  -\cos x}=-\infty$$ because $e^x  -\cos x<0$ for $x\rightarrow0^-$.
Indeed, let $t=-x$. 
Thus, $t\rightarrow0^+$ and
$$e^x-\cos{x}=\frac{1}{e^t}-\cos{t}=\frac{1}{e^t}-1+1-\cos{t}<$$
$$<\frac{1}{1+t}-1+2\sin^2\frac{t}{2}<-\frac{t}{1+t}+\frac{t^2}{2}=\frac{t(t^2+t-2)}{2(1+t)}<0$$
for $t\rightarrow0^+$.
In math $\infty$ it's $+\infty$ or $-\infty$ or like in our case $+\infty$ and $-\infty$.
Thus, the right answer you can see in the first line.
A: 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
$$\frac{\sin x}{e^x -1 -\sin x}\simeq_0\frac x{\dfrac{x^2}2}=\frac 2x\to\begin{cases}+\infty&\text{if }x\to 0^+\\ -\infty&\text{if }x\to 0^-
\end{cases}$$
A: Let's consider the function $f(x)=e^x-\cos x$, whose derivative is $f'(x)=e^x+\sin x$. Since $f'(0)=1>0$, there exists $\delta>0$ such that $f$ is increasing in $[-\delta,\delta]$; in particular
\begin{align}
e^x-\cos x &< 0 && \text{for $-\delta<x<0$} \\[6px]
e^x-\cos x &> 0 && \text{for $0<x<\delta$}
\end{align}
Therefore
$$
\lim_{x\to0^-}\frac{\cos x}{e^x-\cos x}=-\infty
\qquad
\lim_{x\to0^+}\frac{\cos x}{e^x-\cos x}=\infty
$$
By applying l'Hôpital,
$$
\lim_{x\to0^-}\frac{\sin x}{e^x-1-\sin x}=-\infty
\qquad
\lim_{x\to0^+}\frac{\sin x}{e^x-1-\sin x}=\infty
$$
A: 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}
A: Divide the numerator and denominator by $\sin x$ to get $$\lim_{x\to 0}\dfrac{1}{\dfrac{e^{x}-1}{x}\cdot\dfrac{x}{\sin x} - 1}$$ The denominator clearly tends to $0$ so we know that limit does not exist. To further check whether the limit diverges to $\infty$ or $-\infty$ we need to get the sign of denominator for for $x>0$ and $x<0$. This is tricky and does require you to analyze the behavior of $f(x) =e^{x} - 1-\sin x$ using derivatives. One can see that $f'(0)=0,f''(0)=1>0$ so that $x=0$ is a local strict minimum of the function $f$. This proves that $f(x) >0$ in a small deleted neighborhood of $0$.
It follows that $\sin x/(e^{x} - 1-\sin x) $ is negative as $x\to 0^{-}$ and is positive if $x\to 0^{+}$. Therefore the given expression under limit tends to $\infty$ as $x\to 0^{+}$ and tends to $-\infty $ as $x\to 0^{-}$. The limit as $x\to 0$ does not exist (and we say that the given function oscillates infinitely as $x\to 0$).
