$\lim \frac{\cos{x}}{x^2}$ as x goes to infinity $\displaystyle \lim_{x \to \infty} \frac{\cos{x}}{x^2} =\lim_{x \to \infty} \frac{\frac{d}{dx} \cos{x}}{\frac{d}{dx}x^2} = -\frac{1}{2}\lim_{x \to \infty}\frac{\sin{x}}{x}$. But $-\frac{1}{x} \le \frac{\sin{x}}{x} \le \frac{1}{x}$ so $\displaystyle -\lim_{x \to \infty}\frac{1}{x} \le \lim_{x \to \infty}\frac{\sin{x}}{x} \le \lim_{x \to \infty}\frac{1}{x} \iff 0 \le \lim_{x \to \infty}\frac{\sin{x}}{x} \le 0 \iff \lim_{x \to \infty}\frac{\sin{x}}{x} = 0.$
Therefore $\displaystyle \lim_{x \to \infty} \frac{\cos{x}}{x^2} = 0.$
Is the above correct?  
 A: 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 $$\frac{\mid\cos x\mid}{x^2} \leq \frac{1}{x^2} \xrightarrow{\: n \to \infty \: } 0$$
A: 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.
A: $$
\left|\frac{\cos x}{x^2}\right|\le \frac{1}{x^2}\to 0.
$$
A: L'hospital rule is not correct in this case as noticed by @addy2012.
You may observe that you could conclude directly, as $x \to \infty$, by the squeeze theorem:
$$
\left|\frac{\cos x}{x^2}\right|\leq \frac1{x^2}.
$$
A: You don’t need to use the derivatives: For every $x > 0$ we have
$$
 -\frac{1}{x^2} \leq \frac{\cos(x)}{x^2} \leq \frac{1}{x^2}.
$$
Because $\lim_{x \to \infty} \frac{1}{x^2} = 0$ it follows that $\lim_{x \to \infty} \frac{\cos(x)}{x^2} = 0$.
A: Contrary to many other answers (including the accepted answer by adjan) your method is correct. L'Hospital's Rule is applicable for indeterminate forms like $0/0$ and "$\text{anything}/\pm\infty$".
However the use of L'Hospital's Rule is totally unnecessary here and the limit is evaluated more easily by applying "Squeeze theorem" on the inequality $$-\frac{1}{x^{2}}\leq \frac{\cos x}{x^{2}} \leq \frac{1}{x^{2}}$$
