How to prove that $lim_{xtoinfty}frac{(log_2 x)^3}{x^n}=0$

If $K,L$ are positive and $B>1$ then, with $C=1/ln B$, we have (for $x>1$)$$frac {(log_Bx)^K}{x^L}=frac {(Cln x)^K}{x^L}=frac {(Cln x)^K}{(x^{L/K})^K}=$$ $$=frac {(,C(K/L)ln (x^{L/K}),)^K}{(x^{L/K})^K}=$$ $$=C^K(K/L)^Kleft(frac {ln (x^{L/K})}{x^{L/K}}right)^K.$$

Here is a direct way without L'Hospital:

Substituting $x=e^t$ and using $log_2 x= frac{ln x}{ln 2}$ you have

$$lim_{xto infty} frac{(log_2 x)^3}{x^n} =frac 1{ln^3 2}lim_{tto infty}frac{t^3}{e^{nt}}$$

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,

$$0leq frac{t^3}{e^{nt}} < frac{t^3}{frac{(nt)^4}{4!}}=frac{4!}{n^4}cdot frac 1t stackrel{ttoinfty}{longrightarrow}0$$

HINT

$$lim_{xrightarrowinfty}frac{left({log}_2{x}right)^3}{x^n}=lim_{xrightarrowinfty}frac{tleft(ln{x}right)^3}{x^n}$$ where $t=ln^32$ $$lim_{xrightarrowinfty}frac{tleft(1-frac1xright)^3}{x^n}lelim_{xrightarrowinfty}frac{tleft(ln xright)^3}{x^n}lelim_{xrightarrowinfty}frac{tleft(x-1right)^3}{x^n}$$

$$to0lelim_{xrightarrowinfty}frac{tleft(ln xright)^3}{x^n}leto0$$

Now the limit is obvious from squeeze theorem.

For the inequality, refer this

Here's a convincing graph: