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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

So really I am wondering, where do I start with this?

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1 Answer 1

up vote 3 down vote accepted

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?

2) Derivatives may help you find the limits because of the L'Hopital rule. But for this question, you do not need L'Hopital.

3) Yes, that is correct. You need to prove however that the limit is indeed $0$.

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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 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 at 20:40

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