$\lim \frac{\cos{x}}{x^2}$ as x goes to infinity

You must not apply L'Hospital's rule in this case! L'Hospital is only applicable for limits with $\infty \over \infty$ or $0 \over 0$ expressions.

Nonetheless, you can easily get to the limit knowing that $\mid\cos x \mid \leq 1$, thus $$\frac{\mid\cos x\mid}{x^2} \leq \frac{1}{x^2} \xrightarrow{\: n \to \infty \: } 0$$

You arrived at the correct answer, but your first step is incorrect.

Here is another method:

Note that $\cos(x)$ is bounded and $$\lim_{x\rightarrow \infty}\frac{1}{x^2}=0$$ And you're done.

$$ \left|\frac{\cos x}{x^2}\right|\le \frac{1}{x^2}\to 0. $$