Limit of Power of x by Absolute Value of Power of Logarithm of x

Theorem
Let $\alpha$ and $\beta$ be positive real numbers.

Then:


 * $\ds \lim_{x \mathop \to 0^+} x^\alpha \size {\ln x}^\beta = 0$

Proof
From Order of Natural Logarithm Function, we have:


 * $\ln x = \map \OO {x^{-\frac \alpha {2 \beta} } }$ as $x \to 0^+$

That is, by the definition of big-O notation there exists positive real numbers $x_0$ and $C$ such that:


 * $0 \le \size {\ln x} \le C x^{-\frac \alpha {2 \beta} }$

for $0 < x \le x_0$.

So:


 * $0 \le \size {\ln x}^\beta \le C^\beta x^{-\alpha/2}$

for $0 < x \le x_0$.

That is:


 * $0 \le x^\alpha \size {\ln x}^\beta \le C^\beta x^{\alpha/2}$

We have that:


 * $\ds \lim_{x \mathop \to 0^+} C^\beta x^{\alpha/2} = 0$

so by the squeeze theorem for functions:


 * $\ds \lim_{x \mathop \to 0^+} x^\alpha \size {\ln x}^\beta = 0$