How to prove that $\lim_{x\to 0^+} x^p \log x =0$ for any $p>0$? How to prove that $\lim_{x\to 0^+} x^p \log x =0$ for any $p>0$?, without using L'Hôpitals rule or any differentiation or integration?
 A: Using the remark by @julien, we want $\lim_{u\to0^+}u\log u=0$. Let $u=e^{-n}$; then we want $\lim_{n\to\infty}(n/e^n)=0$. So, show that for $n$ sufficiently large, $${n+1\over e^{n+1}}\lt{1\over2}\,{n\over e^n}$$ This will show $n/e^n$ goes to zero faster than $1/2^n$. 
A: Note that if you are allowed to use that $e^v\geq v^2/2$ for $v\geq 0$, which follows readily from the series definition of the exponential, you get a pretty short proof that $v/e^v$ tends to $0$ when $v$ tends to $+\infty$, since
$$
0\leq \frac{v}{e^v}\leq\frac{2v}{v^2}=\frac{2}{v}\qquad\forall v>0.
$$ 
Then $x=e^{-v/p}$ finishes the job.
A: Let $y= \frac{1}{x}$. Then we need to prove that
$$\lim_{y \to \infty}\frac{\log(y)}{y^p}=0$$
To prove this, it suffices to show that
$$\lim_{z \to \infty} \frac{z}{e^{pz}} =0 \,.$$
Let $f(z)=\frac{z}{e^{pz}}$. Then for $z >\frac{1}{p}$ we have
$$f(z+\frac{1}{p})=\frac{z+\frac{1}{p}}{ee^{pz}}<\frac{2}{e}f(z)$$
Then by induction we have 
$$f(z+\frac{n}{p})< \left( \frac{2}{e} \right)^n f(z) \,. (*)$$
Let $M$ be the maximum of $f(z)$ on $[0,\frac{1}{p}]$.
From $(*)$ one can prove by induction that if $\frac{n}{p} \leq z < \frac{n+1}{p}$ we have
$$0< f(z) <  M \left( \frac{2}{e} \right)^n$$
Thus
$$0< f(z) <  M \left( \frac{2}{e} \right)^{\lfloor pz \rfloor}$$
where, $\lfloor pz \rfloor$ denotes the integral part of $pz$. From here, with the squeeze theorem you get the desired result.
A: Let $y = 1/x$.
Then we want to show
$\lim_{y \to \infty} (1/y)^p \ln (1/y) = 0$
or
$\lim_{y \to \infty} \frac{\ln y}{y^p} = 0
$.
For $p=1$,
since $\frac{\ln y}{y}$
has a maximum at $y = e$ and is decreasing
for $y > e$ (via its derivative)
$\frac{\ln y}{y} < 1$ for $y > e$.
Therefore
$\frac{\ln y}{y^2} < \frac1{y}$ for $y > e$,
so 
$\lim_{y \to \infty} \frac{\ln y}{y^2} = 0
$.
This means that,
for any $\epsilon > 0$
we can find a $y_{\epsilon}$ such that
$\frac{\ln y}{y^2} < \epsilon$ for
$y > y_{\epsilon}$.
(From $\frac{\ln y}{y} < 1$ for $y > e$,
we can choose $y_{\epsilon} = \max(e, 1/y)$,
but this is not necessary for what follows, merely useful.)
We want to change the $2$ in $y^2$ to a $p$
so we can get the result for $y^p$.
To do this, let $y = z^{p/2}$.
Then,
for any $\epsilon > 0$
we can find a $y_{\epsilon}$ such that
$\frac{\ln (z^{p/2})}{(z^{p/2})^2} < \epsilon$ for
$z^{p/2} > y_{\epsilon}$,
or
$\frac{(p/2)\ln (z)}{z^p} < \epsilon$ for
$z > (y_{\epsilon})^{2/p}$,
or
$\frac{\ln (z)}{z^p} < (2/p)\epsilon$ for
$z > (y_{\epsilon})^{2/p}$.
Replacing $(2/p)\epsilon$ with
$\epsilon^*$
and
$(y_{\epsilon})^{2/p}$ with
$y^*_{\epsilon}$,
we get
$\frac{\ln (z)}{z^p} < \epsilon^*$ for
$z > y^*_{\epsilon}$.
Given an $\epsilon^*$,
compute $\epsilon = (p/2)\epsilon^*$
and $y^*_{\epsilon} =(y_{\epsilon})^{2/p}$.
Since we have shown that $y_{\epsilon}$ exists,
so does $y^*_{\epsilon}$, and we are done.
Of course, once you are more comfortable with
limits, you can can skip these details
with hand-waving,
but I decided to show in detail
how to go from  the particular exponent of $2$
to the general exponent of $p$.
