Evaluating $\lim\limits_{x\rightarrow 0}\frac{\sin(x)}{\exp(x)-\exp(\sin(x))}$?

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?

• I got $+\infty$ only. Aug 11, 2021 at 16:20
• Since both terms in denominator tend to $1$ you can replace each term with its logarithm. Thus you need to evaluate the limit of $\sin x/(x-\sin x)$. Aug 12, 2021 at 20:51

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,

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

You don't really need to factor like that. You can just say $$\sin(x)=x-x^3/6+o(x^3)$$ so

$$\exp(\sin(x))=1+\sin(x)+\sin(x)^2/2+\sin(x)^3/6+o(\sin(x)^3) \\ = 1 + (x-x^3/6) + (x-x^3/6)^2/2 + (x-x^3/6)^3/6 + o(x^3) \\ = 1 + x + x^2/2 - x^3/6 + x^3/6 + o(x^3) \\ = 1 + x + x^2/2 + o(x^3)$$

so $$\exp(x)-\exp(\sin(x))=x^3/6 + o(x^3)$$.

Your method is a little bit easier in this particular problem but it is pretty limited in its scope compared to this method.

The one actual issue in your calculation is that $$\sin(x)/x \to 1$$ from either side and $$1/x^2 \to +\infty$$ from either side, so you have $$+\infty$$ on both sides.

• Do you mean $\sin(x)=x-x^3/3!+o(x^4)$? Aug 11, 2021 at 16:34
• @KentaS That's true too, but not needed here.
– Ian
Aug 11, 2021 at 16:34
• Oh sorry, I was mixing up big $O$ and little $o$. Aug 11, 2021 at 16:35

We have

\begin{align} &\lim_{x\to 0} \frac{\sin(x)}{\exp(x)-\exp(\sin(x))} =\lim_{x\to 0}\frac{\sin(x)}{x}\cdot\frac{1}{\exp(\sin(x))}\cdot \frac{x}{\exp(x-\sin(x))-1}\\ &=\lim_{x\to 0}\frac{x}{\exp(x-\sin(x))-1}\stackrel{\text{L'Hopital}}=\lim_{x\to 0}\frac{1}{\exp(x-\sin(x))(1-\cos(x))}\\ &=\lim_{x\to 0} \frac{1}{1-\cos(x)} = +\infty \end{align}

As stated in the comments, your solution is correct up until the last step where the limit should be $$+\infty$$.

Since $$f(x)=x−\sin x$$ is increasing in a neighborhood of $$0$$ we can invert this function and use $$\lim _{ x\rightarrow 0 }\frac { e^x -1 }{ x } =1$$ to form $$\lim _{ x\to 0 } \frac { x-\sin x }{ e^x -e^{ \sin x } } =\lim _{ x\to 0 } \frac {1}{ \dfrac{{ e^{ \sin x } \left( e^{ x-\sin x }-1 \right) }}{x-\sin x} } =\lim _{ x\rightarrow 0 }{ \frac{1}{e^{ \sin x }} } =1 .$$ Then since $$\lim _{ x\rightarrow 0 }\frac { x}{ \sin x } =1$$ $$\lim _{ x\to 0 }\frac { \sin x }{x-\sin x}=\lim _{ x\to 0 }\frac{1}{\frac{x}{\sin x}-1}=+\infty.$$

Therefore $$\lim _{ x\to 0 } \frac { \sin x }{ e^x -e^{ \sin x } }=\left(\lim _{ x\to 0 }\frac { \sin x }{x-\sin x}\right)\left(\lim _{ x\to 0 }\frac { x-\sin x }{ e^x -e^{ \sin x } }\right)=+\infty.$$