Prove $\lim_{x \to 0} \frac{e^{\sin(x)} - e^{\tan (x)}}{e^{\sin (2x)}-e^{\tan (2x)}} = \frac{1}{8}$ 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
 A: When $x$ is small $e^x \approx 1+x$ and $\sin(x) \approx x, \tan(x) \approx x$.
Using the above approximations, the limit becomes 
$$\lim_{x \to 0} \frac{\sin(x)-\tan(x)}{\sin(2x)-\tan(2x)}$$ which will be same as
$$\lim_{x \to 0} \frac{\sec(x)-1}{2(\sec(2x)-1)}$$
$$=\lim_{x \to 0} \frac{1-\cos(x)}{2(1-\cos(2x))}$$
$$=\lim_{x \to 0} \frac{x^2/4}{2(x^2)}$$
$$=1/8$$
A: $$F=\lim_{x \to 0} \dfrac{e^{\sin(x)} - e^{\tan (x)}}{e^{\sin (2x)}-e^{\tan (2x)}}$$
$$=\lim_{x\to0}\dfrac{e^{\tan x}}{e^{\tan2x}}\cdot\lim_{x\to0}\dfrac{e^{\sin x-\tan x}-1}{\sin x-\tan x}\cdot\dfrac1{\lim_{x\to0}\dfrac{e^{\sin2x-\tan2x}-1}{\sin2x-\tan2 x}}\cdot\lim_{x\to0}\frac{\sin x-\tan x}{\sin2x-\tan2x}$$
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 $2a=b=1$
