Is $ x \log x = O(x^{1+\epsilon})$ for every $\epsilon > 0$? I am an amateur.
Claim
$$ x \log x = O(x^{1+\epsilon}) \qquad (A) 
$$ 
for every $\epsilon > 0$, $x \in \mathbb{R} \;, x > 2$.
Tried to disproof this, but doubt the proof is correct.
Basic idea:  Define $f(x,\epsilon)= \frac{x \log x}{x^{1+\epsilon}}$,
show that $f(x,\epsilon)$ is unbounded.
I suppose if $a(x)/b(x,\epsilon)$ is unbounded it is impossible
$a(x)=O(b(x,\epsilon))$
Solving $x \log x = x^{1+\epsilon} $ gives 
$\epsilon_0 = -{\frac {\log  \left( x \right) -\log  \left( x\log  \left( x \right)  \right) }{\log  \left( x \right) }}$.
$\epsilon_0$ is positive for $x$ large enough.
Consider 
$$f(x,\epsilon / 10) = {\frac {{x}^{1/10}\log  \left( x \right) }{ \left( x\log  \left( x
 \right)  \right) ^{1/10}}} \qquad (B) $$
(B) is unbounded and the limit at infinity is infinity for positive epsilon. 
I suppose this shows (A) is false.

Is this correct?  In particular "Basic idea" in case of typos.
A correct proof or disproof of (A)?

 A: Basically, you want to know if, when $\alpha>0$,
$$\lim_{x\to\infty} \frac{\log x}{x^\alpha} = 0$$
It's true.
Let $x^\alpha=y$, or $x=y^{1/\alpha}$. Then the limit is the same as
$$\lim_{y\to\infty} \frac{\log \left(y^{1/\alpha} \right)}{y} = \frac 1 \alpha\lim_{y\to\infty} \frac{\log y }{y} $$
Now, you have only one case to check. Let $f(x)=\frac{\log{x}}{x}$, on $[1, +\infty[$, then
$$f'(x)=\frac{1-\log(x)}{x^2}$$
So $f' \leq 0$ and $f\geq 0$, thus $f$ has a limit when $x\to+\infty$.
If this limit is $A$, we already know that $\lim_{x\to\infty} \frac{\log x}{x^\alpha}=\frac{A}{\alpha}$
Then
$$\frac{\log x}{x}=\frac{1}{\sqrt{x}}\frac{\log x}{\sqrt{x}}$$
The first factor tends to $0$ when $x\to+\infty$, and the second to $2A$.
Hence, for any $\alpha>0$,
$$\lim_{x\to\infty} \frac{\log x}{x} =\lim_{x\to\infty} \frac{\log x}{x^\alpha}=0$$
And you can conclude that
$$\log x = o(x^{\alpha}) = O(x^{\alpha})$$
So
$$x \log x = o(x^{1+\alpha}) = O(x^{1+\alpha})$$
The little-o is slightly stronger here, but, it adds not much.

Note: you could of course have differentiated directly 
$$\left(\frac{\log x}{x^\alpha}\right)' = \frac{1-\alpha \log x}{x^{\alpha+1}}$$
And you would finish the proof as easily.
A: It's slightly more efficient to use l'Hopital's rule:
$$\lim_{n\to\infty} \frac{\log n}{n^\alpha}=\lim_{n\to\infty} \frac{1/n}{\alpha n^{\alpha-1}}=\lim_{n\to\infty} \frac{1}{\alpha n^\alpha}=0.$$
