How to show that $\lim_{x \to 0} x^p (\ln x)^r = 0$ I want to show that $\lim_{x \to 0} x^p (\ln x)^r = 0$ if $p > 1$ and $r \in \mathbb{N}$. 
To show this, I wanted to use that $\lim_{x \to 0} x \ln x = 0$, and in fact if $p \geq r$ we can write that 
$$\lim_{x \to 0} x^p (\ln x)^r = \lim_{x \to 0} x^{p-r}((x \ln x)^r)$$
and both terms of the product goes to 0 as $x \to 0$. Thus this part is okay.
I have trouble if $p < r$. I tried to do the same, but I have no result so far. Maybe by considering $(x \ln x)^p (\ln x)^{r-p}$ and then setting $y = \ln x$. Thus we obtain
$$ 
\begin{align*} 
\lim_{x \to 0} (x \ln x)^p (\ln x)^{r-p} 
&= \lim_{e^y \to 0} (e^y \cdot y)^p \cdot y^{r-p} \\
&= \lim_{y \to - \infty} (e^y \cdot y)^p \cdot y^{r-p} \\
&= \lim_{y \to - \infty} e^{py} y^r\\
&= 0
\end{align*}
$$
because the exponential goes "faster" to $0$ than a polynom diverges to infinity? Does it seem correct to you?

Edit:
I will assume that we know that for $r \in \mathbb{N}$, 
$$\lim_{y \to \infty} \frac{y^r}{e^y} = 0.$$
Step 1: I will prove that for $p > 1$ and $r \in \mathbb{N}$ (for my exercise), 
$$\lim_{x \to \infty} \frac{\ln^r x}{x^p} = 0.$$
Proof: Let $y = \ln x$. Then $x = e^y$ and thus
$$
\begin{align*}
\lim_{x \to \infty} \frac{\ln^r x}{x^p} 
&= \lim_{e^y \to \infty} \frac{y^r}{e^{py}} & e^{px} \geq e^x \text{ as } p> 1 \text{ and } x \geq 0\\
&\leq \lim_{y \to \infty} \frac{y^r}{e^y}\\
&= 0.
\end{align*}
$$
Step 2: Now I can prove that under the same hypothesis for $p$ and $r$, 
$$\lim_{x \to 0} x^p \ln^r x = 0.$$
Proof: Let $y = \frac{1}{x}$, thus $x = \frac{1}{y}$ and
$$
\begin{align*}
\lim_{x \to 0} x^p \ln^r x &= \lim_{\frac1y \to 0} \frac{1}{y^p} \cdot \ln \left( \frac{1}{y} \right)^r\\
&= \lim_{y \to \infty} \frac{- \ln^r y}{y^p}\\
&= 0,
\end{align*}
$$
which concludes.
I took $p > 1$ and $r \in \mathbb{N}$ because it is only what I needed for my exercise, but if you feel like editing and proving this for all $p. r > 0$, go ahead.
 A: First let us examine the following. Given \begin{equation} l > 0 \end{equation} we prove that
\begin{equation}
\lim_{x \to0} ln(x)x^l = 0.
\end{equation}
Doing this comes straight from L'Hospitals' rule. We put the limit in the indeterminate form and get the following.
\begin{equation}
\lim_{x \to \infty}\frac{ln(1/x)}{x^l}= \lim_{x \to \infty}\frac{-1}{x} *\frac{1}{lx^{l-1}}=\lim_{x \to \infty} \frac{-1}{lx^{l}} = 0
\end{equation}
Since \begin{equation} l>0 \end{equation}
We now have a very general statement. From here you can see that if you substitute l for p/r (p and r >0), you can put our new found limit to the power of r and still get 0 as your answer. More importantly, by distributivity, we get the wanted limit like so.
\begin{equation}
0=\lim_{x \to 0}(x^{\frac{p}{r}}ln(x))^r =\lim_{x→0}x^p(ln(x))^r
\end{equation}
Hope this helps.
A: I think you would need to prove that $$\lim_{y \to - \infty} e^{py} y^r
= 0$$ using Lo'Hopital's Rule. 
But you could just as easily start the problem by using Lo'Hopital's Rule, by first writing the expression as $$\lim_{x \to 0} \frac{ (\ln x)^r}{x^{-p}}.$$ You would have to use Lo'Hopital's Rule $r$ times, but at the end you should get $$\lim_{x \to 0} \frac{ c}{x^{-p}}$$ where $c$ is a constant. Clearly, this last limit evaluates to $0$. 
