$$\lim_{x \to \infty} \frac{x}{(\ln x)^3} = \infty$$
One way to think of this problem is in terms of the relative growth rates between the numerator and denominator. I know that $x$ grows asymptotically faster than $(\ln x)^3$ according to WolframAlpha. How can I prove this?
$ \lim_{x \to \infty} \frac{x}{(\ln x)^3}=\quad\quad (\frac{\infty}{\infty}\textrm {form, using L'Hospital rule})\\ =\lim_{x \to \infty}\frac{x}{3(\ln x)^2}\quad\quad (\frac{\infty}{\infty}\textrm {form, using L'Hospital rule})\\=\lim_{x \to \infty}\frac{x}{6(\ln x)}\quad\quad (\frac{\infty}{\infty}\textrm {form, using L'Hospital rule})\\ =\lim_{x \to \infty}\frac{x}{6}=\infty$
Put $y = \ln x \implies x = e^y\implies L = \displaystyle \lim_{y \to \infty} \dfrac{e^y}{y^3}$. Can you do Lhospitale three times? or you can use Lhospitale rule directly from the original form above.
Hint. One may observe that, as $x \to \infty$, $$ x=e^{\ln x}>1+\ln x+\frac1{2!}(\ln x)^2+\frac1{3!}(\ln x)^3+\frac1{4!}(\ln x)^4 $$giving, as $x \to \infty$,$$ \frac{x}{(\ln x)^3}>\frac{1}{(\ln x)^3}+\frac{1}{(\ln x)^2}+\frac{1}{2\cdot(\ln x)}+\frac{1}{3!}+\frac{1}{4!}\cdot\ln x $$ and the latter tends to $\infty$.