<s>[INST] 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. 
 [/INST] $ (\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 
$$\lim\limits_{x\to0}\left( (\csc x)^2-\frac{1}{x^2}\right)=\lim\limits_{x\to0}\frac{2x-2\sin x\cos x}{4x^3}=\lim\limits_{x\to0}\frac{2x-\sin 2x}{4x^3}=\lim\limits_{x\to0}\frac{2-2\cos 2x}{12x^2}=\lim\limits_{x\to0}\frac{0+4\sin 2x}{24x}=\lim\limits_{x\to0}\frac{0+8\cos 2x}{24}=\frac13.$$
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