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

Question

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.

Answer 1

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

Answer 2

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

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

Answer 3

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.

Answer 4

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

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

Answer 5

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

Answer 6

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

Answer 7

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

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

Answer 8

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 $\dfrac12$.

Answer 9

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.

Answer 10

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.

Answer 11

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

Answer 12

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)} $

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