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?

As noticed in the comments and given answers, in general, we can’t take the limit only for a single part of the expression with some exceptions as discussed in detail here:

*

*Analyzing limits problem Calculus (tell me where I'm wrong).
Indeed in this case, proceeding by Taylor’s expansion we have
$$\sin^2 x=\left(x-\frac{x^3}6+o(x^3)\right)^2=x^2-\frac{x^4}{3}+o(x^4)$$
pointing out the others terms which play a role in the determining the value for the limit.
