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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?

Any advice will be appreciated.

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    $\begingroup$ You don’t need “$x\neq0$” under your limit. That’s pretty strange in my opinion. $\endgroup$ – gen-z ready to perish Dec 26 '17 at 5:17
  • $\begingroup$ 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. $\endgroup$ – user21820 Dec 26 '17 at 6:29

12 Answers 12

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

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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 $\dfrac12$.

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

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

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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

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

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    $\begingroup$ You are taking the limit inside the series, but that is not always possible, maybe you could justify that step better. $\endgroup$ – Matías Bruna Dec 25 '17 at 23:39
  • $\begingroup$ Thanks, I detailed that step. $\endgroup$ – Zsombor Dec 26 '17 at 8:55
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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

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

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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.

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

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

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

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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.

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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)} $

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

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    $\begingroup$ For another fun limit that is also trivial using this method of asymptotic expansion see here. $\endgroup$ – user21820 Dec 26 '17 at 6:25
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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.

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