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?
 A: Perhaps 
$$\lim_{x\to
 +\infty}\frac{\frac{-1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}=-\frac14\lim_{x\to
 +\infty}\left(\frac{\frac{2}{x}}{\sin\left(\frac{2}{x}\right)}\right)^2=-\frac14$$
A: 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}
A: Let $t=\frac{1}{x},$ then
\begin{equation*}
\lim_{x\rightarrow +\infty }\frac{-\frac{1}{x^{2}}}{\sin ^{2}\left( \frac{2}{%
x}\right) }=\lim_{t\rightarrow 0^{+}}\frac{-t^{2}}{\sin ^{2}\left( 2t\right) 
}=\frac{-1}{4}\lim_{t\rightarrow 0^{+}}\left( \frac{(2t)}{\sin \left(
2t\right) }\right) ^{2}=\frac{-1}{4}\left( \lim_{u\rightarrow 0^{+}}\frac{u}{%
\sin u}\right) ^{2}=-\frac{1}{4}\cdot 1=-\frac{1}{4}.
\end{equation*}
One can use L'Hospital's rule to verify that
\begin{equation*}
\lim_{u\rightarrow 0^{+}}\frac{u}{\sin u}=\lim_{u\rightarrow 0^{+}}\frac{1}{%
\cos u}=\frac{1}{\cos 0}=1.
\end{equation*}
A: $$\lim_{x\to\infty}\frac{-\frac{1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}=$$
$$-\left(\lim_{x\to\infty}\frac{\frac{1}{x^2}}{\sin^2\left(\frac{2}{x}\right)}\right)=$$
$$-\left(\lim_{x\to\infty}\frac{\frac{\text{d}}{\text{d}x}\left(\frac{1}{x^2}\right)}{\frac{\text{d}}{\text{d}x}\left(\sin^2\left(\frac{2}{x}\right)\right)}\right)=$$
$$-\left(\lim_{x\to\infty}\frac{-\frac{2}{x^3}}{-\frac{4\sin\left(\frac{2}{x}\right)\cos\left(\frac{2}{x}\right)}{x^2}}\right)=$$
$$-\left(\lim_{x\to\infty}\frac{\csc\left(\frac{2}{x}\right)\sec\left(\frac{2}{x}\right)}{2x}\right)=$$
$$-\left(\lim_{x\to\infty}\frac{1}{2x\cos\left(\frac{2}{x}\right)\sin\left(\frac{2}{x}\right)}\right)=$$
$$-\frac{1}{2}\left(\lim_{x\to\infty}\frac{1}{x\sin\left(\frac{2}{x}\right)}\right)\left(\lim_{x\to\infty}\frac{1}{\cos\left(\frac{2}{x}\right)}\right)=$$
$$-\frac{1}{2}\left(\lim_{x\to\infty}\frac{1}{x\sin\left(\frac{2}{x}\right)}\right)\left(\frac{1}{\lim_{x\to\infty}\cos\left(\frac{2}{x}\right)}\right)=$$
$$-\frac{1}{2}\left(\lim_{x\to\infty}\frac{1}{x\sin\left(\frac{2}{x}\right)}\right)\left(\frac{1}{\cos\left(\lim_{x\to\infty}\frac{2}{x}\right)}\right)=$$
$$-\frac{1}{2}\left(\lim_{x\to\infty}\frac{1}{x\sin\left(\frac{2}{x}\right)}\right)=-\frac{1}{2}\cdot\frac{1}{2}=-\frac{1}{4}$$
