TITLE: If $f(x)\to +\infty$ as $x\to +\infty$, then why is $\lim_{x\to \infty}\frac{\sin{(x^2+x+1)}}{f(x)}=0$?
QUESTION [7 upvotes]: If $f(x)\to +\infty$ as $x\to +\infty$, then $$\frac{\sin{(x^2+x+1)}}{f(x)}\to 0, \qquad \text{ as } x\to+\infty$$
I know the following is true by the Squeeze Theorem. I am just not sure how to apply it. Any suggestions?

REPLY [2 votes]: Use the fact that sine function is bounded
$$-1\leq \sin x \leq 1$$
$$-1 \leq \sin(x^2+x+1) \leq 1$$
$$\dfrac{-1}{f(x)} \leq \dfrac{\sin(x^2+x+1)}{f(x)} \leq \frac{1}{f(x)}$$
$f(x)→ ∞$ as $x→ \infty $
$$\lim_{x \rightarrow \infty} \dfrac{-1}{f(x)} \leq \lim_{x \rightarrow \infty} \dfrac{\sin(x^2+x+1)}{f(x)} \leq \lim_{x \rightarrow \infty} \frac{1}{f(x)}$$
$$0\leq \dfrac{\sin(x^2+x+1)}{f(x)} \leq 0$$
$$\lim_{x \rightarrow \infty} \dfrac{\sin(x^2+x+1)}{f(x)}=0$$