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

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?

Answer 1

Hint: $-1\le \sin{(x^2+x+1)}\le 1$ for any value of $x$. Hence $$-\frac{1}{f(x)}\le \frac{\sin{(x^2+x+1)}}{f(x)}\le \frac{1}{f(x)}$$

Answer 2

Well, we have $$0\leq|\sin(x^2+x+1)/f(x)|\leq\frac{1}{|f(x)|}$$ Then apply Squeeze Theorem

Answer 3

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 $

$$\lim_{x \rightarrow \infty} \dfrac{-1}{f(x)} \leq \lim_{x \rightarrow \infty} \dfrac{\sin(x^2+x+1)}{f(x)} \leq \lim_{x \rightarrow \infty} \frac{1}{f(x)}$$

$$0\leq \dfrac{\sin(x^2+x+1)}{f(x)} \leq 0$$

$$\lim_{x \rightarrow \infty} \dfrac{\sin(x^2+x+1)}{f(x)}=0$$