# Application of the squeeze theorem

I really just need a bump in the right direction, I am not here looking for an answer as that will not help me at all in the exam.

Use the squeeze theorem to determine.

$\displaystyle \lim \limits_{x \to \infty} \frac{3 - \sin(e^x)}{\sqrt{x^2 + 2}}$

I have no idea how to find this limit, but this is what I do know.

1. My first instinct is that it will have something to do with the limit of $-1 \le \sin(x) \le 1$
2. If I was asked to find the derivative I'm sure that I will get the answer correct, is there anything regarding derivatives that will help me find limits?
3. Just looking at the formula I know the limit is 0

1) Yes, your instincts are right. For the squeeze theorem, you need to find an upper bound and a lower bound for the function $\frac{3-\sin (e^{x})}{\sqrt{x^2+2}}$ so that both of these bounds converge to the same limit. Since $\sin (e^{x}) \geq -1$ for every $x$, one upper bound is $\frac{4}{\sqrt{x^2+2}}$. Now, does this upper bound converge to something, and if it does to what number it converges? Using a similar idea, can you find a lower bound that converges to the same number?
3) Yes, that is correct. You need to prove however that the limit is indeed $0$.
• Ok, so $sin(e^x) \le 1$ which would make the lower bound $\frac{2}{\sqrt{x^2 + 2}}$ it's the $e$ that's putting me off, but I suppose it should not change the bound on $sin$ – Leon Mar 9 '14 at 20:38
• @Leon Yes, the bound $\sin x \leq 1$ holds for every real number $x$. Therefore, $\sin e^x \leq 1$, $\sin(\cos(e^x)+2^{x^2}) \leq 1$, "whatever real number" you put in place of $x$, the bound holds. – Lord Soth Mar 9 '14 at 20:40