$\underset{x\to 0}{\lim }\frac{\tan x-x}{x^2\tan x}$ equals to

$\underset{x\to 0}{\lim }\frac{\tan x-x}{x^2\tan x}=\underset{x\to 0}{\lim }\frac{\sin x-x\cos x}{x^2\sin x}\left(\frac{0}{0}\text{ form }\right)$

$=\underset{x\to 0}{\lim }\frac{\cos x-\cos x+x\sin x}{2x\sin x+x^2\cos x}$

                  $\text{ (using L'Hôpital Rule) }$

$=\underset{x\to 0}{\lim }\frac{\sin x}{2\sin x+x\cos x}=\underset{x\to 0}{\lim }\frac{\left(\sin x\right)/x}{\left(2\right(\sin x)/x)+\cos x}=\frac{1}{3}$