Find limit by l'hospital rule How to calculate,
$$\lim_{x\to 0} \frac{x-\sin x}{x-\tan x} $$
Using L'Hospitals rule.
From original post: 
Find limit by l'hospital rule
$$\lim_{x\to 0} [(x-\sin x)/(x-\tan x)]$$ Where I am wrong? 
$$\lim_{x\to 0} \dfrac{x-\sin x}{x-\tan x}= \lim_{x\to 0} \frac{1-\cos x}{1-(1/\cos^2 x)}=\lim_{x\to 0}\frac{1+\sin x}{1-2\cos^{-3} \sin x}...$$ Now what?

$$\lim_{x\to0}\frac{x-\sin x}{x-\tan x}=\lim_{x\to0}\frac{1-\cos x}{1-\sec^2 x}=-\lim_{x\to0}\frac{\sin x}{2\sec^2 x\tan x}=-\frac{1}{2\sec^3 0}=-\frac12$$
