Show that $\forall \epsilon >0 $ $\exists M >0 $ that $\forall x>M $ $\ln x < x^{\epsilon}$ I thought about defining a function $ g(x) =  \dfrac{1}{\epsilon}  x^{\epsilon} - \ln(x)   $ and show that it's larger than $0$. I saw that if I substitute
$1$, I get $\dfrac{1}{\epsilon}$ which is positive and so on. I also tried to derive the function and I got $g'(x) = x^{\epsilon -1} - \dfrac{1}{x}  $
I'm stuck. Can someone help me realize how to keep going?
edit: I need to show also that $\forall \epsilon >0 $ $\exists M >0 $  that $\forall x>M  $    $\ln x < x^{\epsilon}$ I thought about using what we proved just now by multiplying by $\epsilon$ both sides and get $    \epsilon \ln x < x^{\epsilon}$ define a new function $f(x) =     x^{\epsilon} - \epsilon \ln x$ and again if I derieve it I get $\epsilon x^{\epsilon -1} - \epsilon \dfrac{1}{x}$ get the $\epsilon$ out and get $\epsilon * (x^{\epsilon -1} - \dfrac{1}{x}) > 0 $.
The only problem with it I never use the $M$ can someone help me please?  
Thanks in advance!
 A: If you are allowed to use :
$$\lim_{x\to+\infty}\frac{\ln(x)}x=0$$
then, given $\epsilon>0$, you can see that :
$$\forall x>0,\,\frac{\ln(x)}{x^\epsilon}=\frac 1\epsilon\frac{\ln(x^\epsilon)}{x^\epsilon}$$and since $\lim_{x\to+\infty}x^\epsilon=+\infty$, composition of limits shows that :
$$\lim_{x\to+\infty}\frac{\ln(x)}{x^\epsilon}=0$$
Hence, for $x$ sufficiently large, we certainly have :
$$\frac{\ln(x)}{x^\epsilon}<1$$
A: Let $c > 0$ and $\epsilon \in ]0,1[$. We start with the trivial inequality :
$$
\begin{align}
c &< c + \frac{1}{\epsilon} &\Longleftrightarrow \\
c & < \frac{1}{\epsilon}(1 + \epsilon c) & \overset{(*)}{\implies} \\
c & < \frac{1}{\epsilon} e^{\epsilon c} & \overset{(**)}{\implies} \\
\ln(x) &< \frac{1}{\epsilon} e^{\epsilon \ln(x)} = \frac{1}{\epsilon} x^{\epsilon}
\end{align}
$$
$(*)$ since $1+x \leq e^x, \forall x$.
$(**)$ Setting $c = \ln(x)$ for $x > 1$
A: Alternative you could use L'Hôpital's rule to show
$$\lim_{x\to\infty} \frac{\ln(x)}{x^{\epsilon}} = \lim_{x\to\infty} \frac{x^{-1}}{\epsilon x^{\epsilon-1}} = \lim_{x\to\infty} \frac{1}{\epsilon} x^{-\epsilon} = 0.$$
From this follows that from some point $x^\epsilon > \ln(x)$ otherwise the limit can't be $0$.
Second approach:
Use the $\exp$ function you know that $\exp$ is monoton so the inequality must hold if you add on both sides the exp function
\begin{align*}
\ln(x) < x^\epsilon \quad &\Leftrightarrow \quad \exp(\ln(x)) < \exp(x^\epsilon) \\ &\Leftrightarrow \quad 1 < \frac{\exp(x^\epsilon)}{x} = \sum_{n=0}^\infty \frac{x^{\epsilon n - 1}}{n!} =: f(x)
\end{align*}
So there is only left to show that $f(x) > 1$. For example use $x=\lceil\frac{1}{\epsilon}\rceil!$
 and show that $f$ is monoton from this point or at least still bigger than $1$. Showing that it is still bigger than $1$ should be easy.
