<s>[INST] $\lim \frac{\cos{x}}{x^2}$ 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?  
 [/INST] 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 $\lim_{x \to \infty} \frac{\cos(x)}{x^2} = 0$.
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