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

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:

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

• Start with what you should know like $-1\le \sin(x) \le 1$ Jul 19, 2022 at 0:00
• 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.) Jul 19, 2022 at 0:09
• @RobertShore Oh sorry, I wasn't thinking. $e^{\sin(x)}$ is actually indeterminate, so we can't use L'Hospital here Jul 19, 2022 at 0:14
• What two numbers can you squeeze it between. Jul 19, 2022 at 0:15
• +1 for your try. Jul 19, 2022 at 15:39

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