Calculating $\lim_{x\to\infty} \frac{x^{5}}{2^{\sqrt{x}}}$ I'm interested in calculating
$$\lim_{x\to\infty}\frac{x^{5}}{2^{\sqrt{x}}}$$
I'm thinking of using L'Hopital's Rule here, which gives:
$$\lim_{x\to\infty}\frac{5x^{4}\sqrt{x}}{\ln(2)\cdot
2^{\sqrt{x}-1}}$$
But it doesn't look so great, is this the wrong approach?

This is essentially boils down to making the right substitutions. First we try $x \rightarrow x^2$ to get rid of the square root. Now we have $\frac{x^{10}}{2^x}.$ At this point, you can finish the proof by using the fact that exponentials always outpace polynomials.
Your job is rigorize my 2 statements to an acceptable level. If this is enough for you, then that's fine. If not, I suggest providing a proof for the last step. Either way, good luck!
Note: Now that I think about it, you can prove that $\forall a, n > 1$, $\frac{x^n}{a^x}$ tends to $0$ by strong induction and L'Hopitals rule. 
