Is it true that for all $k>1$, $ x^k > x \log x $ for sufficiently large $x$? I was doing some problems in Big-O notation, and I got intrigued by this question as it was never really addressed. If we plot $x^2$, it's obviously bigger than $x\log x$ for all $x > 0$. However, I started plotting graphs for smaller and smaller values of the exponent, and saw that the crossing point between $x^k$ and $x\log x$ got further and further away. I wondered if maybe there was some particular constant which was the smallest value for $k$ such that the two graphs would cross. But then I found out that even $x^{1.1} >x\log x$ when $x> 3.43 \times 10^{+15}$. This has lead me to believe that for any $k>1$, $x^k$ will eventually overtake $x\log x$, but I think it would be beyond my capabilities to prove it, and I don't know where to start. Does anyone know the answer to this, and is the proof understandable for an undergrad math student?
Edit: Also the functions cross at two points, and I noticed that the first crossing point converges to $e$ as $k$ goes to $1$. What's the explanation for this?
 A: If $k > 1$, then $k = 1 + \varepsilon$ for some $\varepsilon > 0$. So $x^k = x^{1 + \varepsilon} = xx^\varepsilon$.
Thus, to show $x^k = xx^\varepsilon > x\log(x)$, eventually, amounts to showing $x^\varepsilon > \log(x)$, eventually. And, this latter fact is true because:
\begin{align*}
\lim_{x \to \infty}\frac{\log(x)}{x^\varepsilon} &= \lim_{x \to \infty} \frac{1/x}{\varepsilon x^{\varepsilon - 1}}\quad \text{ (L'Hôpital's rule)} \\
&= \lim_{x \to \infty} \frac{1}{\varepsilon x^\varepsilon} = 0\quad \text{ ($x^\varepsilon \to \infty$ as $\varepsilon > 0$)}
\end{align*}
Answer to Edit:
The reason you have a crossing point converging to $e$ as $k \to 1$ is quite simple. Just let $k = 1$ and let us calculate the crossing points. Crossing points occur when: $x^k = x = x \log(x)$. Or rearranging:
$$
x - x\log(x) = x (1 - \log(x)) = 0
$$ If you ignore $x = 0$, then the above is true when $1 - \log(x) = 0$ or $\log(x) = 1$ or $x = e$.
A: Write
$$x^k=x \log(x)\implies x^{k-1}=\log(x)\implies x^{k-1}=\frac 1{k-1}\log \left(x^{k-1}\right)$$
Let $y=x^{k-1}$ to make
$$y=\frac 1{k-1}\log(y)\implies y=\frac{W(1-k)}{1-k}$$ where $W(.)$ is Lambert function.
In the real domain, this function is defined as long as $1-k \geq -\frac 1e$ that is to say $k \leq 1+\frac 1e$.
Back to $x$, the  curves intersect at
$$x=\left(\frac{W(1-k)}{1-k}\right)^{\frac{1}{k-1}}$$ which is an increasing function.
When $k\to 1$, $x\to e$ and when $k\to  1+\frac 1e$, $x\to e^e$.
We can expand $x$ as series. For example, when $k$ is close to $1^+$, we have
$$x=e\left(1+(k-1)+2 (k-1)^2+\frac{13}{3} (k-1)^3+\frac{235}{24} (k-1)^4+O\left((k-1)^5\right) \right)$$ Try it for $k=\frac 54$; the approximation will give $x=\frac{3033 e}{2048}\approx 4.02566$ while the exact value is $4.17708$<.
A: Yes, it is  a basic result in asymptotic analysis that  for any $\alpha,\beta >0$,
$$\ln^\alpha x=_{\infty}o\bigl(x^\beta\bigr). $$
A: The statement is equivalent to
$$
x^\varepsilon   > \log x,
$$
for any $\varepsilon>0$ and sufficiently large $x$. From the power series expansion of the exponential function $e^x>x$ for all $x>0$, i.e., $x>\log x$ for all $x>0$. Thus,
$$
\log x = \frac{2}{\varepsilon }\log (x^{\varepsilon /2})  < \frac{2}{\varepsilon }x^{\varepsilon /2}  < x^{\varepsilon /2} x^{\varepsilon /2}  = x^\varepsilon  ,
$$
whenever $x^{\varepsilon /2}>2/\varepsilon \Leftrightarrow x > (2/\varepsilon )^{2/\varepsilon }$.
