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

Question

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

Answer 1

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, $$ \lim_{x\to0^-}\frac{\sin x}{e^x-1-\sin x}=-\infty \qquad \lim_{x\to0^+}\frac{\sin x}{e^x-1-\sin x}=\infty $$

Answer 2

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

Answer 3

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}

Answer 4

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.

Answer 5

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

Answer 6

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 $x\to 0$).