<s>[INST] Finding $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$ 
What is the limit of $\,\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right)$ ?

My thought was $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\left(\frac{1}{\sin^2 x} - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\left(\frac{1}{x^2} - \frac{1}{x^2} \right) = 0$. I used $\lim\limits_{x\to 0} \frac{\sin x}{x} = 1$.
But $\lim\limits_{x\to 0}\,\left(\csc^2x - \frac{1}{x^2}\right) = \lim\limits_{x\to 0}\,\frac{x^2-\sin^2x}{x^2\sin^2x}$ and if I apply L'Hospital's Rule four times at the second expression, I get $\lim\limits_{x\to 0}\,\left(\frac{8\cos 2x}{24\cos 2x - 32\sin 2x - 8x^2\cos 2x}\right) = \frac {1}{3}$.
What am I missing?
 [/INST] Without L'Hopital, we can rewrite the limmand as
$$\frac{x-\sin x}{x^3}\cdot\frac{x+\sin x}{x}$$
The limit of the product is the product of the limits when both exist. The term on the right goes to $2$, so all there is left is to evaluate the term on the left. By using the identity
$$\sin 3x = 3\sin x - 4\sin^3x$$
notice that
$$L = \lim_{x\to0}\frac{x-\sin x}{x^3} = \lim_{u\to0}\frac{3u-\sin 3u}{(3u)^3} = \lim_{u\to 0}\frac{3u - 3\sin u +4\sin^3 u}{27u^3}$$
$$ = \lim_{u\to 0}\frac{u-\sin u}{9u^3}+\frac{4\sin^3u}{27u^3} = \frac{1}{9}L+\frac{4}{27} \implies L = \frac{1}{6}$$
Therefore the original limit is $\frac{1}{6}\times 2 = \boxed{\frac{1}{3}}$
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