$\mathop{\lim}\limits_{x\to+\infty} \dfrac{\ln x}{x^\varepsilon}, \varepsilon>0$ without L'Hospital's rule I got stuck when proving $\ln x = o\left( {{x^\varepsilon }} \right)$ ($\varepsilon>0$). 
Any ideas without L'Hospital's rule?
$$\mathop{\lim}\limits_{x\to+\infty} \dfrac{\ln x}{x^\varepsilon}, \varepsilon>0$$
Thanks in advance.
 A: We could show the following
\begin{align}
\ln x \leq x^{\varepsilon/2}
\end{align}
for $x\ge 1$. Consider
\begin{align}
h(x) = x^{\varepsilon/2}-\ln x
\end{align}
where $h(1) = 1>0$ and $h'(x) = \frac{\epsilon x^{\epsilon/2}}{2x}-\frac{1}{x} = \frac{1}{x}\left(\frac{\epsilon}{2}x^{\epsilon/2}-1 \right)>0$
when $x>1$.  Hence it follows
\begin{align}
h(x)\geq 0 \ \ \Rightarrow \ \ x^{\varepsilon/2}\geq \ln x. 
\end{align}
Finally, we have that
\begin{align}
\left|\frac{\ln x}{x^\epsilon} \right|\leq \frac{x^{\epsilon/2}}{x^{\epsilon}}\leq \frac{1}{x^{\epsilon/2}} \rightarrow 0
\end{align}
as $x\rightarrow \infty$. 
A: Your problem is equivalent to showing $\ln x=o(x)$, since $\ln x/x\le \delta $ for $x\ge x_0$ if and only if $\ln x/x^{\epsilon}\le \delta/\epsilon$ for $x\ge (x_0)^{1/\epsilon}$.
Furthermore, by similar reasoning, showing $\ln x/x\to 0$ is equivalent to proving $x/e^x\to 0$. 
Let $n(x)$, which I will just write as $n$, be the largest integer less than $x$, so $x-1\le n\le x$. Using the binomial theorem,
$$
e^x\ge e^n=(1+(e-1))^n\ge\binom{n}{2}(e-1)^2\ge \frac12 (x-1)(x-2)(e-1)^2,
$$
so 
$$
\frac{x}{e^x}\le \frac{x}{(x-1)(x-2)(e-1)^2}\to 0,
$$
proving $x/e^x\to0$.
A: $\mathop{\lim}\limits_{x\to+\infty} \dfrac{\ln x}{x^\varepsilon}, \varepsilon>0
$
For a first step,
since
$\dfrac{\ln x}{x^\varepsilon}
=\dfrac1{\varepsilon}\dfrac{\ln x^{\varepsilon}}{x^\varepsilon}
$,
$\dfrac{\ln x}{x^\varepsilon}
\to 0
$
if and only if
$\dfrac{\ln x}{x}
\to 0
$.
So we only need to prove that
$\dfrac{\ln x}{x}
\to 0
$.
If $f(x) =  \dfrac{\ln x}{x}$,
$f'(x)
=\dfrac{1-\ln x}{x^2}
< 0
$
for
$x > e$.
$f(x^2)
=\dfrac{\ln x^2}{x^2}
=\dfrac{2\ln x}{x^2}
=\dfrac{\ln x}{x}\dfrac{2}{x}
$,
so that
$\dfrac{f(x^2)}{f(x)}
=\dfrac{2}{x}
$.
Therefore,
if $x > 2e$,
$\dfrac{f(x^2)}{f(x)}
<c
$
where
$c = \dfrac{2}{e}
< 1$.
By an easy induction,
if $x > 2e$,
$\dfrac{f(x^{2^k})}{f(x)}
<c^k
$
or
$f(x^{2^k})
< f(x) c^k
$.
Since
$c^k \to 0$,
for any
$\epsilon > 0$
and
for any $x > 2e$
we can choose
$k$ large enough
so that
$f(x^{2^k})
< \epsilon$.
Therefore
$\lim_{x \to \infty}f(x)
= 0$.
