Prove $2^{\sqrt{\log n}}=o(n)$ To my understanding, I need to show the following equals 0.
I tried using L'hopital's rule, but got the same $\lim$ times a constant.
$$
\lim_{n \to \infty} \frac{2^{\sqrt{\log_e n}}}{n}
$$
 A: Il will consider the logarithm is base $e$, but it doesn't change anything if you consider another base.
You have to prove that :

$$\lim_{n\to\infty}\frac{2^\sqrt{\ln(n)}}n=0$$

Take its logarithm :
$$\ln\left(\frac{2^\sqrt{\ln(n)}}n\right)=\ln(2)\sqrt{\ln(n)}-\ln(n)=\ln(n)\left(\frac{\ln(2)}{\sqrt{\ln(n)}}-1\right)$$
It is clear that the last expression  in parenthesis has limit $-1$, hence :
$$\lim_{n\to\infty}\ln\left(\frac{2^\sqrt{\ln(n)}}n\right)=-\infty$$
and the conclusion follows, by composition of limits.
A: We will use the fact that
\begin{align}
\sqrt{\log n} \leq \frac{3}{4}\log n
\end{align}
when $n$ is sufficiently large. Then it follows
\begin{align}
e^{\sqrt{\log n}} \leq e^{\frac{3}{2}\log\sqrt{n}} = n^{3/4}.
\end{align}
Hence the rest follows immediately.
A: L'Hôpital: $$\lim_{n \to \infty} \frac{2^{\sqrt{\log n}}}{n} = \frac{\log(2)}{2} \lim_{n \to \infty} \frac{ 2^{\sqrt{\log n}}}{n \sqrt{\log (n)}}$$
Therefore if $\lim_{n \to \infty}\frac{2^{\sqrt{\log n}}}{n} = A$ where $A < \infty$, then $A = \frac{\log(2)}{2} \times 0$, since the limit on the right-hand is $0$ by "limit of a product is the product of the limits".
So the limit you seek is either $0$ or $\infty$. That's as far as I quickly see how to go with L'Hôpital.

Note that the expression is $$2^{\sqrt{\log n} - \log(n)/\log(2)}$$
so the limit you seek is $$\exp \left(\log(2) \lim_{n \to \infty} \left[\sqrt{\log n} - \frac{\log n}{\log 2}\right] \right)$$
Substituting $u=\log n$, we need to find $$L = \lim_{u \to \infty} \left[\sqrt{u} - \frac{u}{\log 2}\right]$$
But differentiation shows that the expression in the brackets is negative for $u > \log(2)^2$ and gets more negative approximately linearly, so $L$ must be $-\infty$ and hence your original limit is $0$. (To be more formal, one can argue that your original limit is either $0$ or $\infty$; it can't be $\infty$ because $L \leq 0$, so it must be $0$.)
