Prove that $\frac{1}{x^{1+\epsilon}}<\frac{1}{x(\log x)^p}$ 
Given $p>0$, $\epsilon>0$, prove that $\displaystyle \frac{1}{x^{1+\epsilon}}<\frac{1}{x(\log x)^p}$ for sufficiently large $x$.

If $p\leq \epsilon$, then $(\log x)^p\leq x^p\leq x^\epsilon$. So the result is clear. But what if $p>\epsilon$? Any suggestion please.
 A: It's enough to show that $(\log x)^p<x^{\varepsilon}$ for all sufficiently large $x$, or equivalently $\log x<x^{\frac{\varepsilon}{p}}$ for all $x$.
To show this, observe that for any $a>0$ we have
$$ \lim_{x\to\infty}\frac{\log x}{x^a}=\lim_{x\to\infty}\frac{1}{ax^a}=0 $$
hence $\log x<x^a$ for all sufficiently large $x$. Now take $a=\frac{\varepsilon}{p}$.
A: HINT:
The inequality is equivalent to the inequality 
$$1<\frac{x^\epsilon}{\log^p(x)}$$
for sufficiently large $x$
Now, evaluate the limit
$$\lim_{x\to \infty}\frac{x^\epsilon}{\log^p(x)}$$
SPOILER ALERT:  Scroll over the highlighted area to reveal the solution

Note that we have $$\lim_{x\to \infty}\frac{x^\epsilon}{\log^p(x)}=\infty$$Therefore, for any number $B>0$, there exists a number $L$ such that whenever $x>L$, $$\frac{x^\epsilon}{\log^p(x)}>B \tag 1$$Noting that $B$ can be any positive number, we choose $B=1$.  Thus, there exists an $L>0$ large enough so that whenever $x>L$, $$\frac{x^\epsilon}{\log^p(x)}>1$$  

A: Check with l'Hospital the limit
$$\lim_{x\to\infty}\frac{(\log x)^p}{x^\epsilon}=\lim_{x\to\infty}\frac{p(\log x)^{p-1}}{\epsilon x^\epsilon}=\lim_{x\to\infty}\frac{p(p-1)(\log x)^{p-2}}{\epsilon^2 x^\epsilon}=\ldots$$
You can see that at some point the power in numerator will be less than that of $\;x\;$ in the denominator...
A: Note that we have for any $\gamma>0$, 
$$\log(x)<\frac{x^\gamma}{\gamma} \tag 1$$
In particular, $(1)$ is true when $\gamma <\epsilon/p$.  
Then, we have
$$\begin{align}
\frac{x^\epsilon}{\log^p(x)}&>\frac{\gamma x^\epsilon}{x^{p\gamma}}\\\\
&=\gamma x^{\epsilon-p\gamma}\\\\
&>1
\end{align}$$
when $x>\left(\frac{1}{\gamma}\right)^{1/(\epsilon-p\gamma)}$.
For example, $(1)$ is true for $\gamma =\frac{\epsilon}{2p}$.  Thus, we find that 
$$\frac{1}{x^{1+\epsilon}}<\frac{1}{x\log^p(x)}$$
whenever $x>\left(\frac{2p}{\epsilon}\right)^{2/\epsilon}$
A: Many ways lead to Rome. Some are longer than others. If I am not mistaking the question is to deduce (in a simple way)  from $\log x < x$ for $x$ large that for any $0<a<1$ we have $\log x < x^a$ for $x$ large enough ($a\geq 1$ has already been taken care of)?
If this is the case note that $\log x = \frac{2}{a}\log(x^{a/2}) < \frac{2}{a}x^{a/2}\leq   x^{a/2} x^{a/2}=x^a$ for $x$ large (since$\frac{2}{a}\leq x^{a/2}$ for $x$ large).
