# Determine $\lim\limits_{x\to 0, x\neq 0}\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{1}{2}$ without using L'Hospital

How to prove that

$$\lim\limits_{x\to 0, x\neq 0}\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{1}{2}$$

without using L'Hospital?

Using L'Hospital, it's quite easy. But without, I don't get this. I tried different approaches, for example writing $$e^{\sin(x)}=\sum\limits_{k=0}^\infty\frac{\sin(x)^k}{k!}$$ and $$\sin(2x)=2\sin(x)\cos(x)$$ and get $$\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{\sin(x)+\sum\limits_{k=2}^\infty\frac{\sin(x)^k}{k!} }{2\sin(x)\cos(x)}$$ but it seems to be unrewarding. How can I calculate the limit instead?

• You don’t need “$x\neq0$” under your limit. That’s pretty strange in my opinion. Dec 26, 2017 at 5:17
• In most cases, L'Hopital's rule is inferior to asymptotic expansion. Not only does it sometimes fail to lead anywhere, even when it works it is much more cumbersome, whereas asymptotic expansion can often be done mentally very quickly, as you can see from my answer. Dec 26, 2017 at 6:29

From the known limit $$\lim\limits_{u\to 0}\frac{e^u-1}{u}=1,$$ one gets $$\lim\limits_{x\to 0}\frac{e^{\sin x}-1}{\sin(2x)}=\lim\limits_{x\to 0}\left(\frac{e^{\sin x}-1}{\sin x}\cdot\frac{\sin x}{\sin(2x)}\right)=\lim\limits_{x\to 0}\left(\frac{e^{\sin x}-1}{\sin x}\cdot\frac{1}{2\cos x}\right)=\color{red}{1}\cdot\frac{1}{2}.$$

It's much simpler; rewrite your faction as $$\frac{\mathrm e^{\sin x}-1}{2\sin x\cos x}=\frac{\mathrm e^{\sin x}-1}{\sin x}\,\frac1{2\cos x}.$$ Setting $u=\sin x$, the first fraction is the rate of variation of $\mathrm e^u$: $$\frac{\mathrm e^{\sin x}-1}{\sin x}=\frac{\mathrm e^u-1}{u} \xrightarrow[\,u\to 0\,]{}1,\quad \frac{1}{2\cos x}\xrightarrow[\,x\to 0\,]{}\frac1{2}$$ hence the limit is $\dfrac12$.

$$\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{e^{\sin x}-1}{\sin x}\frac{\sin x}{x}\frac{2x}{\sin 2x}\frac12=1\cdot1\cdot1\cdot\frac12=\frac{1}{2}$$

Using the Taylor polynomial, for the exponential, and cancelling $\sin x$ on the last step, $$\frac{e^{\sin x}-1}{\sin 2x}=\frac{1+\sin x+o(\sin^2x)-1}{2\sin x\cos x} =\frac{1+o(\sin x)}{2\cos x}\xrightarrow[x\to 0]{}\frac12.$$

From your last equation, for nonzero $\sin 2 x$

$$\frac{\sin(x)+\sum\limits_{k=2}^\infty\frac{\sin(x)^k}{k!} }{2\sin(x)\cos(x)}= \frac{1+\sum\limits_{k=2}^\infty\frac{\sin(x)^{k-1}}{k!} }{2\cos(x)} \longrightarrow \frac{1}{2}$$ because $$\Bigg\vert \sum\limits_{k=2}^\infty\frac{\sin(x)^{k-1}}{k!} \Bigg\vert \leq \vert \sin(x) \vert \sum\limits_{k=2}^\infty\frac{\vert\sin(x)^{k-2}\vert}{k!} \leq \vert \sin(x)\vert \sum\limits_{k=2}^\infty\frac{1}{k!}\leq \vert \sin(x) \vert\longrightarrow 0$$ since

$$x\rightarrow 0,\quad \sin(x)\rightarrow 0,\quad \cos(x)\rightarrow 1$$

• You are taking the limit inside the series, but that is not always possible, maybe you could justify that step better.
– Nah
Dec 25, 2017 at 23:39
• Thanks, I detailed that step. Dec 26, 2017 at 8:55

Hint Let $f(x)=e^{sin(x)}-1$. Then

$$\lim_{x \to 0} \frac{f(x)-f(0)}{x-0}=f'(0)$$

Also, $$\lim_{x \to 0} \frac{x}{\sin(2x)}$$ can be easily be deduced from the fundamental trigonometric limit.

Alternately canceling $\sin(x)$ you get

$$\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{\sin(x)+\sum\limits_{k=2}^\infty\frac{\sin(x)^k}{k!} }{2\sin(x)\cos(x)}=\frac{1+\sum\limits_{k=2}^\infty\frac{\sin(x)^{k-1}}{k!} }{2\cos(x)}$$

Hint: If I continue with your approach $$\frac{e^{\sin(x)}-1}{\sin(2x)}=\frac{\sin(x)+\sum\limits_{k=2}^\infty\frac{\sin(x)^k}{k!} }{2\sin(x)\cos(x)}=\frac{1+\sum\limits_{k=2}^\infty\frac{\sin(x)^{k-1}}{k!} }{2\cos(x)}$$

You can now set $x=0$, but you'll need to justify this step.

As $x\to0$, we can consider the fact that for $x\approx 0$, we have $\sin x\approx \tan x\approx x$

\begin{align}L&=\lim_\limits{x\to0}\dfrac{e^{\sin x}-1}{\sin 2x}\\&=\lim_\limits{x\to0}\dfrac{e^x-1}{2x}\\&=\lim_\limits{x\to 0}\dfrac{1+x-1}{2x}\qquad[\because \text{For }x\approx 0,e^x\approx1+x]\\&=\lim_\limits{x\to0}\dfrac{x}{2x}\\&=\dfrac12\end{align}

$$\lim_{x\to 0}\frac{e^{\sin(x)}-1}{\sin(2x)}=\lim_{x\to 0}\frac{\sin(x)+\sum\limits_{k=2}^\infty\frac{\sin(x)^k}{k!} }{2\sin(x)\cos(x)}$$ $$=\lim_{x\to 0}\frac{1+\sum\limits_{k=2}^\infty\frac{\sin(x)^{k-1}}{k!} }{2\cos(x)}$$ $$=\frac{1}{2}$$

The Taylor series for $e^{\sin x}-1$ about $x=0$ is $x +\dfrac{x^2}2-\dfrac{x^4}8+\cdots$ and the Taylor series about $x=0$ for $\sin 2x$ is $2x -\dfrac{8x^3}6+\dfrac{32x^5}{120}+\cdots$. Now we can divide the two series and take limit as $x$ approaches $0$. Factor $x$ from top and bottom and cancel it. The desired limit of $\dfrac12$ is found.

All the existing answers involve some thinking, but actually there is a totally systematic way (that even common computer algebra systems use):

As $x \to 0$:

$2x \to 0$ and $\sin(x) \to 0$.

Thus $\dfrac{e^{\sin(x)}-1}{\sin(2x)} \in \dfrac{(1+\sin(x)+o(\sin(x)))-1}{2x+o(2x)} \subseteq \dfrac{\sin(x)+o(x+o(x))}{2x+o(x)}$

$\quad \subseteq \dfrac{(x+o(x))+o(x)}{2x+o(x)} = \dfrac{1+o(1)}{2+o(1)} \to \dfrac12$.

• For another fun limit that is also trivial using this method of asymptotic expansion see here. Dec 26, 2017 at 6:25

First you prove that, $\lim_{x\to0} \frac{1}{x} ln (1+x) = 1$

$\lim_{x\to0} \frac{1}{x} ln (1+x)$ = $\lim_{x\to0}ln (1+x)^\frac{1}{x}$ = $ln$$(\lim_{x\to0}(1+x)^\frac{1}{x})$ = $ln (e)$ = 1

Now to evaluate $\lim_{x\to0} \frac{e^x-1}{x}$ put, $e^x = 1 + z$ then, $x= ln(1+z)$ and $z\rightarrow{0}$ when $x\rightarrow{0}$ So, $\lim_{x\to0} \frac{e^x-1}{x}$ = $\lim_{x\to0} \frac{z}{ln(1+z)}$ = 1 (by above limit formula for $ln$)

Now you are on the position to understand and use any one of the solutions explained above.