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 all boils down to the proof that exponential functions tend to infinity much faster than polynomials as the argument tends to infinity or equivalently the polynomials tend to infinity much faster than logarithmic functions.
In symbols we have for any $k>0, a>1$ $$\lim_{x\to\infty}\frac{x^{k}}{a^x}=0=\lim_{x\to\infty}\frac{\log x} {x^k}\tag{1}$$ The above results are pretty standard and can be used anywhere just the way we use the more famous limit formula $$\lim_{x\to 0}\frac{\sin x} {x} =1\tag{2}$$ Your question is now easily solved by using the substitution $x=t^2$ to get $$\lim_{x\to \infty} \frac{x^5}{2^{\sqrt{x}}}=\lim_{t\to\infty} \frac{t^{10}}{2^t}=0$$ In case you want a proof of the formula $(1)$ it is an easy consequence of the properties of exponential and logarithmic functions.
We can use the fundamental inequality $\log x\leq x-1$ to get $$\log x=\frac{1}{p}\log x^p\leq \frac{x^p-1}{p}<\frac{x^p}{p}$$ where $p$ is a number such that $0<p<k$ and then for $x>1$ we have $$0<\frac{\log x} {x^k} <\frac{1}{px^{k-p}}$$ and by Squeeze Theorem we get $$\lim_{x\to\infty} \frac{\log x} {x^k} =0$$ For the exponential limit just use the substitution $a^x=t$ so that $x=(\log t) /\log a$ and $$\frac{x^k} {a^x} =(\log a) ^{-k} \cdot\left(\frac{\log t} {t^{1/k}}\right)^{k}\to (\log a) ^{-k} \cdot 0=0$$
