<s>[INST] Limit without hospital's rule $\lim_{x\rightarrow 0} (\frac{1}{x^2} - \frac{x}{\sin^3(x)})$ 
Calcule $\lim_{x\rightarrow 0} (\frac{1}{x^2} - \frac{x}{\sin^3(x)})$ without using L'Hospital

I first was trying to replace $x = \frac{\pi}{2} + u$ but I can't find a rule or eliminate anything.
After that I try using cubes:
$\frac{\sin^3(x) - x^3}{x^2\sin^3(x)} = \frac{(\sin(x)-x)(\sin^2(x)+x^2+x\sin(x))}{x^2\sin^3(x)}$ but again i couldn't find any rule or formula.
How can I solve this problem without using L'Hospital? Thank you very much in advance
 [/INST] $$
\begin{align}
\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{x}{\sin^3x}\right) &= \lim_{x\to 0}\left(\frac{\sin^3x-x^3}{x^2\sin^2x}\right) \\
&= \lim_{x\to 0}\left(\frac{x^3}{\sin^3x}\right)\left(\frac 1{x^2}\right)\left(\frac{\sin^3x}{x^3}-1\right) \\
&=1\cdot\lim_{x\to 0}\left(\frac{1}{x^2}\right)\left(\frac{\sin x}{x}-1\right)\left(\frac{\sin^2x}{x^2}+\frac{\sin x}{x}+1\right) \\
&=1\cdot\lim_{x\to 0}\left(\frac{\sin x-x}{x^3}\right)\cdot3 \\
&=1\cdot\left(\frac {-1}6\right)\cdot3 \\
&=\frac{-1}{2}
\end{align}
$$
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