$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}$ without using L'Hopital $$\lim_{x\to0} \frac{x-\sin x}{x-\tan x}=?$$
I tried using $\lim\limits_{x\to0} \frac{\sin x}x=1$.
But it doesn't work :/

$$\frac{x - \sin(x)}{x - \tan(x)} = \frac{x - \sin(x)}{x^3} \cdot \frac{x^3}{x - \tan(x)}$$
Let $x = 3y$ and $x\to 0 \implies y\to 0$
$$\lim_{x\to0} \frac{x - \sin(x)}{x^3} = L $$
$$L = \lim_{y\to0}\frac{3y - \sin(3y)}{(3y)^3} = \lim_{y\to0} \frac 3 {27} \frac{y - \sin(y)}{y^3} + \lim_{y\to0} \frac{4}{27} \frac{\sin^3(y)}{y^3} = \frac{1}{9} L + \frac 4{27} $$
This gives 
$$\lim_{x\to0}\frac{x - \sin(x)}{x^3} = \frac 1 6 $$
\begin{align*}
L &= \lim_{y\to0}\frac{ 3y - \tan(3y)}{27 y^3} \\ 
 &= \lim_{y\to0} \frac{1}{(1 - 3\tan^2(y ))} \cdot \frac{3y(1 - 3\tan^2(y )) - 3 \tan(y) + \tan^3(y)}{27y^3}\\ 
 &= \lim_{y\to0} \frac{1}{(1 - 3\tan^2(y ))} \cdot \left( 
\frac 3 {27} \frac{y - \tan(y)}{y^3} + \frac  1 {27} \frac{\tan^3(y)}{y^3}  - \frac 9 {27} \frac{y \tan^2(y)}{y^3 }
 \right )\\ 
 &= \frac {3L}{27} + \frac 1 {27} - \frac 1 3 \\ 
\end{align*}
This gives other limit to be $-1/3$, put it up and get your limit.
