2
$\begingroup$

I want to evaluate this limit and I faced with one issue.
for this post I set $L`$ as L'Hôpital's rule $$\lim\limits_{x\to 0}\left(\frac{1+x\sin(2x)-\cos(x)}{\sin^2(x)}\right)$$ Solution One:
$$\lim\limits_{x\to 0}\left(\frac{1+x\sin(2x)-\cos(x)}{\sin^2(x)}\right)=\frac{0}{0}L`=\frac{\sin(2x)+2x\cos(2x)+\sin(x)}{\sin(2x)}$$ at this step I decided to evaluate each fraction so I get :
$$\lim\limits_{x\to 0}\frac{\sin(2x)}{\sin(2x)}+\frac{2x\cos(2x)}{\sin(2x)}+\frac{\sin(x)}{\sin(2x)} = \frac{3}{2}$$

Solution Two:
$$\lim\limits_{x\to 0}\left(\frac{1+x\sin(2x)-\cos(x)}{\sin^2(x)}\right)\frac{0}{0}L`=\frac{\sin(2x)+2x\cos(2x)+\sin(x)}{\sin(2x)}=\frac{0}{0}L`$$ $$\frac{2\cos(2x)+2\cos(2x)-4x\sin(2x)+\cos(x)}{2\cos(2x)}=\frac{5}{2}$$
I would like to get some idea where I did wrong, Thanks.

$\endgroup$
1
$\begingroup$

As mentioned your first solution is incorrect. the reason is $$\displaystyle lim_{x\to0}\frac{2xcos(2x)}{sin(2x)}\neq 0$$ you can activate agin l'hospital: $$lim_{x\to0}\frac{2xcos(2x)}{sin(2x)}=lim_{x\to0}\frac{2cos(2x)-2xsin(2x)}{2cos(2x)}=lim_{x\to0} 1-2xtg(2x)=1+0=1$$, so now after we conclude this limit, in the first solution, i'd write after $$\lim\limits_{x\to 0}\left(\frac{1+x\sin(2x)-\cos(x)}{\sin^2(x)}\right)=\lim\limits_{x\to 0}\frac{\sin(2x)}{\sin(2x)}+\frac{2x\cos(2x)}{\sin(2x)}+\frac{\sin(x)}{\sin(2x)}$$ that the limit equals to $$ = 1+\frac 1 2+1=\frac 5 2$$ and that's the correct answer.

$\endgroup$
0
$\begingroup$

Solution two is correct (the notation is a little strange though).

In your first limit when you say you'll "evaluate each fraction", what is the contribution of the middle fraction? I suspect that's where you've gone awry.

$\endgroup$

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.