1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – Namaste Jan 14 '15 at 17:41
  • $\begingroup$ Sorry, I just improved to what was already there. $\endgroup$ – Aaron Maroja Jan 14 '15 at 17:42
  • $\begingroup$ @Aaron Yes, I see. I directed my comment, now, to sbp. $\endgroup$ – Namaste Jan 14 '15 at 17:43
  • $\begingroup$ @amWhy oh, okay then. $\endgroup$ – Aaron Maroja Jan 14 '15 at 17:44
3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Missing a $-$ sign $\endgroup$ – David Peterson Jan 14 '15 at 17:35
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.