Using l'Hopital's rule to evaulate $\lim\limits_{x\to0}\left( (\csc x)^2-\frac{1}{x^2}\right)$ Use l'Hopital's rule to find the following limit
$$\lim\limits_{x\to0}\left( (\csc x)^2-\frac{1}{x^2}\right)$$
I tried differentiation but did not get the right answer. I know I have to put it into a single fraction. 
 A: Putting into a single fraction
$$
\csc^2(x)-\frac{1}{x^2} = \frac{1}{\sin^2(x)}-\frac{1}{x^2}=\frac{x^2-sin^2(x)}{x^2sin^2(x)}
$$
Try differentiation now. You will get that the limit is $\frac{1}{3}$ as the answer.
A: $$\lim_{x\to 0}\left((\csc x)^2-\frac{1}{x^2}\right)=\lim_{x\to 0}\left(\frac{\frac{x^2}{sin^2(x)}-1}{x^2}\right)$$
We get the form $0/0$, so applying L'Hopital for the first time we get $$\displaystyle \lim_{x\to0}\left((\csc x)^2-\frac{1}{x^2}\right)=\lim_{x\to0}\left(\large \frac{\frac{2x \sin^2(x)-2x^2 \sin(x)\cos(x)}{\sin^4(x)}}{\normalsize 2x}\right)=\lim_{x\to0}\left(\frac{\sin(x)-x \cos(x)}{\sin^3(x)}\right)$$
Again we get the form $0/0$, so applying L'Hopital for the second time we get $$\displaystyle \lim_{x\to0}\left((\csc x)^2-\frac{1}{x^2}\right)=\lim_{x\to0}\left(\frac{\cos(x)-\cos(x)+x \sin(x)}{3\sin^2(x)\cos(x)}\right)= \\ =\frac{1}{3}\lim_{x\to0}\left(\frac{x}{\sin(x)}\right)\cdot\lim_{x\to0}\left({\frac{1}{\cos(x)}}\right)$$
Using the known limit $\displaystyle \lim_{x\to0}\frac{x}{\sin(x)}=1$ we eventually get $$\displaystyle \lim_{x\to0}\left((\csc x)^2-\frac{1}{x^2}\right)=\frac{1}{3}$$
A: $ (\csc x)^2-\frac{1}{x^2}=\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x^2-\sin^2x}{x^2\sin^2x }$ ~ $\frac{x^2-\sin^2x}{x^4 }$
So using  l'Hopital's rule 
$$\lim\limits_{x\to0}\left( (\csc x)^2-\frac{1}{x^2}\right)=\lim\limits_{x\to0}\frac{2x-2\sin x\cos x}{4x^3}=\lim\limits_{x\to0}\frac{2x-\sin 2x}{4x^3}=\lim\limits_{x\to0}\frac{2-2\cos 2x}{12x^2}=\lim\limits_{x\to0}\frac{0+4\sin 2x}{24x}=\lim\limits_{x\to0}\frac{0+8\cos 2x}{24}=\frac13.$$
