<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] $$\lim_{x\to0}\left(\csc^2(x)-\frac{1}{x^2}\right)=\lim_{x\to0}\left(\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}\right)$$
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.
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