# Powers Drown Logarithms

## Theorem

Let $r \in \R_{>0}$ be a (strictly) positive real number.

Then:

$\ds \lim_{x \mathop \to \infty} x^{-r} \ln x = 0$

### Corollary

$\ds \lim_{y \mathop \to 0_+} y^r \ln y = 0$

## Proof

When $x > 1$:

$\forall s \in \R: s > 0: \ln x \le \dfrac {x^s} s$

Given that $r > 0$, we can plug $s = \dfrac r 2$ in:

 $\ds x^{-r} \ln x$ $=$ $\ds x^{-r/2} \paren {x^{-s} \ln x}$ $\ds$ $\le$ $\ds \frac {x^{-r/2} } s$ $\ds$ $=$ $\ds s \frac 1 {x^{r/2} }$
$\ds \lim_{x \mathop \to \infty} x^{-r} \frac 1 {x^{r/2} } = 0$

and so:

$\ds \lim_{x \mathop \to \infty} x^{-r} \ln x = 0$

$\blacksquare$