<s>[INST] Using L'Hopital to solve $\lim_{x\to +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$ 
Use L'Hopital to calculate
$$\lim_{x\to
 +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}$$

Right now this yields $\frac{0}{0}$ so let's go ahead and use L'Hopital:
$$\lim_{x\to
 +\infty}\frac{\frac{2}{x^3}}{2\cdot \sin\left(\frac{2}{x}\right)\cdot\cos \left(\frac{2}{x}\right)\cdot\left(\frac{-2}{x^2}\right)}$$
This just won't do. Perhaps we should flip the numerator with the denominator instead:
$$-\frac{\csc^2\left(\frac{2}{x}\right)}{x^{2}}$$
This yields -$\frac{\infty}{\infty}$, so we can go ahead and apply L'Hopital:
$$-\frac{2\cdot-\csc\left(\frac{2}{x}\right)\cdot\cot\left(\frac{2}{x}\right)\cdot\frac{-2}{x^2}}{2x}$$
If I evaluate this I will get $\frac{0}{\infty}$
I have the feeling I'm not supposed to keep going this path, and there's a simpler solution (using L'Hopital). What can I do to solve this?
 [/INST] Use the substitution $u=\frac{2}{x}$
Then the limit becomes
\begin{align}
\lim_{u\to 0} \frac{-\frac{1}{4}u^2}{\sin^2{u}}&=\lim_{u\to 0}\frac{-\frac{1}{2}u}{2\sin u\cos u}\\ \\
&=-\frac{1}{4}\lim_{u\to 0}\frac{u}{\sin u\cos u}\\ \\
&=-\frac{1}{4}\lim_{u\to 0}\frac{1}{\cos^2 u-\sin^2 u}\\ \\
&=-\frac{1}{4}
\end{align}
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