I've been trying to tackle this limit for a couple of days now to no avail, so I believe the way to solve this is simply beyond my rusty Calc 1 & 2 skills. I've posted the attempt that has gotten me the furthest, to demonstrate my thought process, but it still looks like I'm stuck in a "loop" of sorts.
$\lim_{x \to \infty}\frac{2^\sqrt{log_2x}}{(\log_2x)^2} = \lim_{x \to \infty}\frac{2^\sqrt{log_2x}\ln{2}\frac{\frac{1}{2}(log_2x)^{\frac{-1}{2}}}{x\ln2}}{\frac{2(\log_2x)}{x\ln{2}}} = \lim_{x \to \infty}{\frac{(\ln{2})2^\sqrt{log_2x}}{4(\log_2x)^{\frac{5}{2}}}} \frac{\ln2}{4}=\lim_{x \to \infty}{\frac{2^\sqrt{log_2x}}{(\log_2x)^{\frac{5}{2}}}}$
Viewing the two equations $2^\sqrt{log_2x}$ and $(\log_2x)^2$ on a graph makes me believe the solution is $\infty$, but I won't feel satisfied until I reach a solution.
Any help would be greatly appreciated!
Let $y=\sqrt {\log_2 x}\;.$ Then $y\to \infty$ as $x\to \infty.$ And $$\frac {2^{\sqrt {\log_2 x}}}{(\log_2 x)^2}=\frac {2^y}{y^4}\to \infty \text { as } y\to \infty.$$ For example let $[y]$ be the largest integer not exceeding $y.$ When $y\ge 5$ we have $2^y\geq$ $2^{[y]}=$ $\sum_{j=0}^{[y]}\binom {[y]}{j}>$ $ \binom {[y]}{5}=$ $\frac {1}{5!}\prod_{j=0}^4([y]-j)>$ $ \frac {1}{5!}\prod_{i=1}^5 (y-i).$