Powers Drown Logarithms

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

Then:
 * $\displaystyle \lim_{x \mathop \to \infty} x^{-r} \ln x = 0$

Proof
From Upper Bound of Natural Logarithm:

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:

From Sequence of Powers of Reciprocals is Null Sequence:
 * $\displaystyle \lim_{x \mathop \to \infty} x^{-r} \frac 1 {x^{r/2}} = 0$

and so:
 * $\displaystyle \lim_{x \mathop \to \infty} x^{-r} \ln x = 0$

by the Squeeze Theorem for Real Sequences.