How do I prove that $\lim_{x \rightarrow \infty} \frac{\ln^m x}{x^{1/n}} = 0$ for all m and n I want to prove that
$$\lim_{x \rightarrow \infty} \frac{\ln^m x}{x^{1/n}} = 0$$
for all $m, n \in \mathbb{N}$. I tried using L'Hopital's rule, but that got me nowhere.  
 A: ${{\ln^m(x)}\over {x^{1\over n}}}= \left({\ln(x)\over{x^{1\over{nm}}}} \right)^m$
Write $u=x^{1\over{nm}}$, so that you have \begin{align*}
\lim_{x\rightarrow +\infty}{{\ln(x)}\over x^{1\over {nm}}} &=\lim_{u\rightarrow +\infty}{{\ln(u^{nm})}\over u} \\
&=nm \lim_{u\rightarrow +\infty}{{\ln(u)}\over u} \\ &=0.
\end{align*}
A: HINT: Setting $$t=\ln(x)$$ then we get
$$\frac{t^m}{e^{t}}$$ clerly that the searched Limit is Zero for $t$ tends to infinity
A: Hint:
$$ \frac{\ln^m x}{x^{1/n}} = \frac{(\ln x)^m}{(x^{\frac{1}{nm})^m}} = \left(\frac{\ln x}{x^{\frac{1}{nm}}}\right)^m =\left(nm\frac{\ln x^{\frac{1}{nm}}}{x^{\frac{1}{nm}}}\right)^m  = \left(nm\frac{\ln X }{X}\right)^m $$
With $$X=x^{\frac{1}{nm}}\to\infty~~~as~~x\to\infty $$
i.e $$\lim_{x\to\infty} \frac{\ln^m x}{x^{1/n}} =\lim_{X \to\infty} \left(nm\frac{\ln X }{X}\right)^m  = 0$$
A: $\displaystyle \lim_{x \rightarrow \infty} \frac{\ln^m x}{x^{1/n}} = 0$
$\dfrac{\ln^m x}{x^{1/n}} =\exp\bigg[\ln\bigg(\dfrac{\ln^m x}{x^{1/n}}\bigg)\bigg]=\exp\bigg[\ln(\ln^mx)-\ln x^{\frac{1}{n}}\bigg]=\exp\bigg[m\ln(\ln x)-\dfrac{1}{n}\ln x\bigg]$
$=\exp\bigg[-\dfrac{1}{n}\ln x\bigg(1-\underbrace{mn\dfrac{\ln(\ln x)}{\ln x}}_{\to 0}\;\bigg)\bigg]$
thus  $\lim_\limits{x \rightarrow \infty} \dfrac{\ln^m x}{x^{1/n}} = \lim_\limits{x \rightarrow \infty}\dfrac{1}{x^{1/n}}=0$
A: In
$\lim_{x \rightarrow \infty} \frac{\ln^m x}{x^{1/n}}
$,
set
$x = e^t$
so
$\ln x = t$.
Then
$\frac{\ln^m x}{x^{1/n}}
=\frac{t^m}{e^{t/n}}
=\left(\frac{t^{mn}}{e^{t}}\right)^{1/n}
$.
Since
$e^t
=\sum_{k=0}^{\infty} \dfrac{t^k}{k!}
$,
it $t > 0$
then
$e^t
\gt\dfrac{t^k}{k!}
$
for any $k$,
so
$\dfrac{t^{k+1}}{e^t}
\lt (k+1)!
$
or
$\dfrac{t^{k}}{e^t}
\lt \dfrac{(k+1)!}{t}
$.
Therefore
$\lim_{t \to \infty}\dfrac{t^{k}}{e^t}
=0
$
for any $k$.
Now let $k = mn$.
