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.


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.


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.


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