Limit of $\frac{2^n}{n^{\sqrt{n}}}$

You need some (easy) asymptotic analysis to obtain the limit.

As already noted $$\frac{2^n}{n^{\sqrt{n}}}=\mathrm e^{n\ln 2-\sqrt{n}\ln n}=\mathrm e^{n\ln 2\bigl(1-\tfrac{\ln n}{\sqrt n\ln 2}\bigr)}.$$ Now it is standard that $$\ln n=o(\sqrt n)\enspace\text{ – in other words },\; \lim_{n\to\infty} \frac{\ln n}{\sqrt n}=0.$$ Therefore, the second factor in the exponent tends to $1$, and consequently the exponent itself tends to $+\infty$.