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We want to find the limit of

$$\lim_{x\to \infty} \frac{e^{\sin(x)}}{\ln(\ln(x))}$$

Since $\lim_{x\to\infty}e^{\sin(x)} = \infty$ and $\lim_{x\to\infty}\ln(\ln(x)) = \infty$

I tried using 'Hospital here, which leads to

$$\frac{\mathrm{e}^{\sin\left(x\right)}\cos\left(x\right)}{\dfrac{1}{x\ln\left(x\right)}}$$

but that also cannot be evaluated.

But if we insert the limit of our original function, we get this: enter image description here

Can someone explain, how one can apply the squeeze theorem here or solve this with another approach?

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    $\begingroup$ Start with what you should know like $ -1\le \sin(x) \le 1$ $\endgroup$ Jul 19, 2022 at 0:00
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    $\begingroup$ What makes you think $\lim e^{\sin x} = \infty$? You need that to be true to justify using L'Hopital's Rule, so you need to justify that step. (You'll find that you can't, because it's not true.) $\endgroup$ Jul 19, 2022 at 0:09
  • $\begingroup$ @RobertShore Oh sorry, I wasn't thinking. $e^{\sin(x)}$ is actually indeterminate, so we can't use L'Hospital here $\endgroup$ Jul 19, 2022 at 0:14
  • $\begingroup$ What two numbers can you squeeze it between. $\endgroup$
    – homosapien
    Jul 19, 2022 at 0:15
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    $\begingroup$ +1 for your try. $\endgroup$
    – homosapien
    Jul 19, 2022 at 15:39

1 Answer 1

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Claim: $$\lim_{x \to\infty} \frac{e^{\sin x}}{\ln(\ln(x))}=0.$$

Proof of claim:

Since $\sin x$ has a minimum value of $-1$ and a maximum value of $1$, You have that

$$\frac{1}{e \ln(\ln(x))} \le \frac{e^{\sin x}}{\ln(\ln(x))}\le \frac{e}{\ln(\ln(x))}$$

holds for every $x \to \infty$, forcing the limit to be $0$ To see this note $\frac{1}{e \ln(\ln(x))}, \frac{e}{\ln(\ln(x))}$ both tend to $0$ as $x \to \infty$. $\blacksquare$

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