TITLE: Find limit by l'hospital rule
QUESTION [1 upvotes]: 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?

REPLY [0 votes]: Apply l'Hôpital:
$$
\lim_{x\to0}\frac{x-\sin x}{x-\tan x}=
\lim_{x\to0}\frac{1-\cos x}{1-\dfrac{1}{\cos^2x}}
$$
At this point it's much better to simplify, before proceeding further:
$$
\lim_{x\to0}\frac{\cos^2 x(1-\cos x)}{\cos^2x-1}=
\lim_{x\to0}\frac{\cos^2 x}{-(\cos x+1)}
$$
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