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?
 A: 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 $\frac{1}{6}\times 2 = \boxed{\frac{1}{3}}$
A: $$\lim_{x\to0}\left(\csc^2(x)-\frac{1}{x^2}\right)=\lim_{x\to0}\left(\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}\right)$$
Both numerator and denominator tend to zero, and they are continuous and differentiable, so L'Hopital's rule applies. The derivative of the numerator is: $2x-2\sin(x)\cos(x)=2x-\sin(2x)$. The derivative of the denominator is: $2x\sin^2(x)+2x^2\sin(x)\cos(x)=2x\sin^2(x)+x^2\sin(2x)$. Well, these both will still tend to zero in the limit, so we can simply reapply L'Hopital.
The second derivative of the numerator is: $2-2\cos(2x)$. The second derivative of the denominator is:
$$2\sin^2(x)+4x\sin(x)\cos(x)+2x\sin(2x)+2x^2\cos(2x)\\=2\sin^2(x)+4x\sin(2x)+2x^2\cos(2x)$$ Which still tends to zero.
The third derivative of the numerator is $4\sin(2x)$. The third derivative of the denominator is:
$$4\sin(x)\cos(x)+8x\cos(2x)+4\sin(2x)+4x\cos(2x)-4x^2\sin(2x)\\=6\sin(2x)+12x\cos(2x)-4x^2\sin(2x)$$
Well, it is our misfortune to have to apply this rule four times. The fourth derivative of the numerator is: $8\cos(2x)\to8,x\to0$. The fourth derivative of the denominator is: $12\cos(2x)+12\cos(2x)-24x\sin(2x)-8x\sin(2x)-8x^2\cos(2x)\to24,x\to0$. At  last!
The original limit is the ratio of the (fourth) derivatives of the numerator and denominator, which amounts to $\frac{8}{24}=\frac{1}{3}$.
A quick visual check with Desmos will confirm this, if the visuals help.
A: 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:

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*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.
