Computing $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3}$ without L'Hôpital's rule. 
Computing $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3}$ without L'Hopital

Say $\lim_{x\to{0+}}\frac{\tan(x)-x}{x^3} = L$
For $L$:
$$L=\lim_{x\to0}\frac{\tan x-x}{x^3}\\
L=\lim_{x\to0}\frac{\tan 2x-2x}{8x^3}\\
4L=\lim_{x\to0}\frac{\frac12\tan2x-x}{x^3}\\
3L=\lim_{x\to0}\frac{\frac12\tan{2x}-\tan x}{x^3}\\
=\lim_{x\to0}\frac{\tan x}x\frac{\frac1{1-\tan^2x}-1}{x^2}\\
=\lim_{x\to0}\frac{(\tan x)^3}{x^3}=1\\
\large L=\frac13$$
I found that in another Q, can someone tell me why 
$$L=\lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\frac{\tan 2x-2x}{8x^3}$$

Let $x=\arctan y.$ We have $y\to 0^+$ as $x\to 0^+.$
For $x\in (0,\pi /2)$ we have $\frac {\tan x-x}{x^3}=$ $\frac {y-\arctan y}{y^3}\frac {y^3}{x^3}=$ $\frac {(y-\arctan y)}{y^3}(\frac {\tan x}{x})^3.$
We have $\lim_{x\to 0^+}(\frac {\tan x }{x})^3=1.$ 
Consider $g(y)=y-\arctan y.$
We have $g'(y)-y^2=$ $(1-\frac {1}{1+y^2})-y^2=$ $\frac {-y^4}{1+y^2}\in [-y^4,0].$ 
Therefore $g(y)=g(y)-g(0)=$ $\int_0^y g'(t)dt=$ $y^3/3-\delta \cdot y^5/5$ for some $\delta \in [0,1].$
