Proving that $\ln \ x = o(x^{-p}) \ \ \forall p>0$ as $x \rightarrow 0^{+}$ $$\ln \ x = o(x^{-p}) \ \ \forall p>0$$ as $$x \rightarrow 0^{+}$$
Using the definition, I want to prove that $$\lim_{x \to 0^+} \frac{\ln \ x}{x^{-p}} = 0$$
I see that as $x$ goes to zero, $\ln x$ goes to minus infinity and $x^p$ goes to $0$. Does this mean the multiple will then be zero? How do I make this more rigorous?
Thanks.
 A: Replacing $x=e^{-y}$, it suffices to check that
$$ \lim_{y \to \infty} y e^{-py} = 0. $$
Using L'Hôpital's rule,
$$ \lim_{y \to \infty} \frac{y}{e^{py}} = \lim_{y \to \infty} \frac{1}{pe^{py}} = 0. $$
Alternatively, from the definition of $e^y$ as $\lim_{n \to \infty} (1+y/n)^n$, we have
$$ \log{x} = \lim_{n \to \infty} n(x^{1/n}-1), $$
or if we look at $e^{-y}$ instead,
$$ \log{x} = \lim_{n \to \infty} -n(x^{-1/n}-1), $$
(think of $\log{x}=-\log{(1/x)}$). In particular, using the fact that the first increases and the second decreases in $n$ for fixed $x$, one can check that we have the inequalities
$$ -n(x^{-1/n}-1) \leqslant \log{x} \leqslant m(x^{1/m}-1) $$
for any $m,n>0$. (For $m=n=1$, this inequality reduces to the well-known $(1-x^{-1} \leqslant \log{x} \leqslant x-1)$.
Therefore we find that for $0<x<1$,
$$ 0<-x^p\log{x} < n(x^{p-1/n}-x^p), $$
and for sufficiently large $n$, both of the terms in brackets tend to $0$ as $x \to 0$.
A: Start with the known limit $x\ln x\to 0$ as $x\to 0$. Then write
$$
\frac{\ln x}{x^{-p}}=\frac1p x^p\ln x^p=\frac1p y\ln y
$$
with the monotonically increasing parameter substitution $y=x^p$.
A: One more way: set $\log x = t$, hence $x = e^t$ and $x^p = e^{tp}$. After a bit of algebraic manipulation notice the $0< e^{-1}<1$, so if you set $x=e^{-1}$, the function $f(x) = x^s$ is uniformly continuous. 
$$
L = - \frac{x}{p} \lim_{s \to \infty} \frac{d}{dx}x^s = -\frac{x}{p} \frac{d}{dx} \lim_{x \to \infty} x^s = 0
$$ 
because,as explaineda bove, $0<x<1$. So the limit is $0 \ \forall \ p>0$ 
A: Using L'Hospital, we can show
$$
\begin{align}
\lim_{x\to0^+}x\log(x)
&=\lim_{x\to0^+}\frac{\log(x)}{1/x}\\
&=\lim_{x\to0^+}\frac{1/x}{-1/x^2}\\[6pt]
&=\lim_{x\to0^+}-x\\[9pt]
&=0\tag{1}
\end{align}
$$
Then substituting $x\mapsto x^p$ into $(1)$ yields
$$
\lim_{x\to0^+}x^pp\log(x)=0\tag{2}
$$
which is equivalent to
$$
\lim_{x\to0^+}\frac{\log(x)}{x^{-p}}=0\tag{3}
$$
which says
$$
\log(x)=o(x^{-p})\tag{4}
$$
