TITLE: Prove $\lim_{x \to 0} \frac{e^{\sin(x)} - e^{\tan (x)}}{e^{\sin (2x)}-e^{\tan (2x)}} = \frac{1}{8}$
QUESTION [5 upvotes]: Here's a nice little problem.
$$\lim_{x \to 0} \frac{e^{\sin(x)} - e^{\tan (x)}}{e^{\sin (2x)}-e^{\tan (2x)}}$$
What's the quickest way to do this? One line solutions will be applauded :D
Cheers, my jolly people :D

REPLY [4 votes]: $$F=\lim_{x \to 0} \dfrac{e^{\sin(x)} - e^{\tan (x)}}{e^{\sin (2x)}-e^{\tan (2x)}}$$
$$=\lim_{x\to0}\dfrac{e^{\tan x}}{e^{\tan2x}}\cdot\lim_{x\to0}\dfrac{e^{\sin x-\tan x}-1}{\sin x-\tan x}\cdot\dfrac1{\lim_{x\to0}\dfrac{e^{\sin2x-\tan2x}-1}{\sin2x-\tan2 x}}\cdot\lim_{x\to0}\frac{\sin x-\tan x}{\sin2x-\tan2x}$$
Using $\lim_{h\to0}\dfrac{e^h-1}h=1,$
$$F=\lim_{x\to0}\frac{\sin x-\tan x}{\sin2x-\tan2x}$$
Now for finite $a,b\ne0,$
$$F=\lim_{x\to0}\frac{\sin2ax-\tan2ax}{\sin2bx-\tan2bx}$$
$$=\lim_{x\to0}\frac{\cos2bx}{\cos2ax}\cdot\lim_{x\to0}\frac{1-\cos2ax}{1-\cos2bx}\cdot\lim_{x\to0}\frac{\sin2ax}{\sin2bx}$$
$$=\frac{\cos0}{\cos0}\cdot\lim_{x\to0}\frac{2\sin^2ax}{2\sin^2bx}\cdot\lim_{x\to0}\frac{\sin2ax}{2ax}\cdot\dfrac1{\lim_{x\to0}\dfrac{\sin2bx}{2bx}}\cdot\frac ab$$
$$=\frac{a^2}{b^2}\cdot\left(\lim_{x\to0}\frac{\sin ax}{ax}\right)^2\dfrac1{\left(\lim_{x\to0}\dfrac{\sin bx}{bx}\right)^2}\cdot1\cdot\dfrac11\frac ab$$
$$=\frac{a^3}{b^3}$$
Here $2a=b=1$