How to prove that $\lim_{x\to\infty}\frac{(\log_2 x)^3}{x^n}=0$ 
I need help with proving the following $$\lim_{x\to\infty}\frac{\left({\log}_2x\right)^3}{x^n}=0\ , \quad \forall n>0.$$

I used wolframalpha and got it

and I had kind of an intuition because of the limit $\frac{\ln(x)}x$ , and still I have no idea how to formaly proof it.
Will appreciate any help!
 A: Using the hint of @Kavi Rama Murthy, note that $$f(x)\to 0 \implies f^{3}(x)\to 0$$
In your problem, you can define that $f(x):=\frac{\log_{2}(x)}{x^{n/3}}$ so, by L'Hospital's Rule you can find that $$\frac{\log_{2}(x)}{x^{n/3}} \to 0 \quad \text{as} \quad x \to \infty$$
So, by the hint and since that $f^{3}(x)=\frac{\log_{2}^{3}(x)}{x^{n}}$ so $$\frac{\log_{2}^{3}(x)}{x^{n}} \to 0 \quad \text{as} \quad x\to \infty$$
this's true for all $n>0$.
A: HINT
$$\lim_{x\rightarrow\infty}\frac{\left({\log}_2{x}\right)^3}{x^n}=\lim_{x\rightarrow\infty}\frac{t\left(\ln{x}\right)^3}{x^n}$$
where $t=\ln^32$
$$\lim_{x\rightarrow\infty}\frac{t\left(1-\frac1x\right)^3}{x^n}\le\lim_{x\rightarrow\infty}\frac{t\left(\ln x\right)^3}{x^n}\le\lim_{x\rightarrow\infty}\frac{t\left(x-1\right)^3}{x^n}$$
$$\to0\le\lim_{x\rightarrow\infty}\frac{t\left(\ln x\right)^3}{x^n}\le\to0$$
Now the limit is obvious from squeeze theorem.
For the inequality, refer this
Here's a convincing graph:

A: Here is a direct way without L'Hospital:
Substituting $x=e^t$ and using $\log_2 x= \frac{\ln x}{\ln 2}$ you have
$$\lim_{x\to \infty} \frac{(\log_2 x)^3}{x^n} =\frac 1{\ln^3 2}\lim_{t\to \infty}\frac{t^3}{e^{nt}}$$
Now, since $e^u = \sum_{k=0}^{\infty}\frac{u^k}{k!}$, you have $e^u > \frac{u^4}{4!}$ for any $u>0$. Hence,
$$0\leq \frac{t^3}{e^{nt}} < \frac{t^3}{\frac{(nt)^4}{4!}}=\frac{4!}{n^4}\cdot \frac 1t \stackrel{t\to\infty}{\longrightarrow}0$$
A: If $K,L$ are positive and $B>1$ then, with $C=1/\ln B$, we have (for $x>1$)$$\frac {(\log_Bx)^K}{x^L}=\frac {(C\ln x)^K}{x^L}=\frac {(C\ln x)^K}{(x^{L/K})^K}=$$ $$=\frac {(\,C(K/L)\ln (x^{L/K})\,)^K}{(x^{L/K})^K}=$$ $$=C^K(K/L)^K\left(\frac {\ln (x^{L/K})}{x^{L/K}}\right)^K.$$
