I need some help with my homework. How do I prove that
$$\lim_{n\to\infty}\frac{(\frac{n^3}{\log_{2}n})}{4^\sqrt{n}} =0\:?$$
I need to compare the growth rate of these functions and I know that $4^\sqrt{n}$ grows faster and thats why the limit should be $0$.
But how do I solve this? What steps should I show that lead this equation to being $0$?
Thank you for your help!
Let $n\in\mathbb{N}$. Observe that $n^3\le4^{\sqrt n}$ and from this it follows that $$\frac{(\frac{n^3}{\log_{2}n})}{4^\sqrt{n}} =\frac{n^3}{\log_{2}(n)4^\sqrt{n}}\le \frac{1}{\log_{2}(n)}.$$
If we let $n\to\infty$ we can see that $$\lim_{n\to\infty}\frac{1}{\log_{2}(n)}=0.$$
In conclusion: $$\lim_{n\to\infty}\underbrace{\frac{(\frac{n^3}{\log_{2}n})}{4^\sqrt{n}}}_{\text{non-negative}}\le0\Longrightarrow \lim_{n\to\infty}{\frac{(\frac{n^3}{\log_{2}n})}{4^\sqrt{n}}}=0$$