# 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?

• @sbp: by leaving out the OPs displayed effort, you put this post in jeopardy of being closed and deleted. The OP did indeed show his/her work!! Whoever approved the edit aided in putting this post at risk. – Namaste Jan 14 '15 at 17:41
• Sorry, I just improved to what was already there. – Aaron Maroja Jan 14 '15 at 17:42
• @Aaron Yes, I see. I directed my comment, now, to sbp. – Namaste Jan 14 '15 at 17:43
• @amWhy oh, okay then. – Aaron Maroja Jan 14 '15 at 17:44

## 3 Answers

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

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.

• Missing a $-$ sign – David Peterson Jan 14 '15 at 17:35

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: $$\lim_{x\to0}\frac{\cos^2 x(1-\cos x)}{\cos^2x-1}= \lim_{x\to0}\frac{\cos^2 x}{-(\cos x+1)}$$