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?

When you differentiate the numerator and denominator a second time, you should get $$\lim_{x\to 0} \dfrac{\sin x}{-2\cos^{-3} x\sin x} = \lim_{x\to 0}\frac{\sin x}{-2\sec^2 x \tan x}$$
Note that $\frac{d}{dx}(1) = 0$, so you lose the summands of $1$ in the numerator and denominator.
So you were indeed on the right track...all you need is to make the correction, and apply l'hospital once again.
