Evaluating: $\lim_{x \to 0}\frac{\sin(\tan x)-\tan(\sin x)}{x-\sin x}$ I was asked to evaluate the following limit:
$$\lim_{x \to 0}\frac{\sin(\tan x)-\tan(\sin x)}{x-\sin x}$$
I really have no idea how to solve it!

Using Taylor of oder 4, near $x= 0$ see here  we have 
$$\sin x\sim x-\frac{x^3}{6}+o(x^4)~~~and ~~~~\tan x\sim x +\frac{x^3}{3} +o(x^4)$$ 
but $\tan x\sim x+\frac{x^3}{3}$ stay near 0 as x is near 0 therefore, 
$$\sin(\tan x) \sim \sin(x+\frac{x^3}{3}) \sim x+\frac{x^3}{3}  -\frac{1}{6}(x+\frac{x^3}{3})^3 = x+ \frac{x^3}{6}  +o(x^4)$$
similarly, $\sin x\sim x-\frac{x^3}{6}$ stay near 0 as x is near 0 therfore,
$$\tan(\sin x) \sim  x-\frac{x^3}{6}  +\frac{1}{3}(x-\frac{x^3}{6})^3 \sim x+\frac{x^3}{6} + o(x^4)$$
also 
$$x-\sin x  \sim \frac{1}{6}x^3 +o(x^4)$$
we conclude that 
$$\frac{\sin(\tan x)-\tan(\sin x)}{x-\sin x}\sim \frac{x+\frac{x^3}{6}-x-\frac{x^3}{6} +o(x^4)}{\frac{x^3}{6} +o(x^4) } = \frac{0+o(x)}{\frac{1}{6} +o(x) }$$
that is $$\color{red}{\lim_{x\to 0}\frac{\sin(\tan x)-\tan(\sin x)}{x-\sin x} = 0}.$$
OR By L'hopital rule 
we have 
 $$\lim_{x\to 0}\frac{\sin(\tan x)-\tan(\sin x)}{x-\sin x} \\ = 
\lim_{x\to 0}\frac{(1+\tan^2x)\cos(\tan x)-\cos x(1+\tan^2(\sin x))}{1-\cos x} \sim \frac{0+o(x^2)}{\frac{x^2}{2} +o(x^2)} = 0 $$
by just taking Taylor of oder 2 to get the same answer.
