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

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Theorem

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

Then:

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


Corollary

Let $k$ be a positive real number.

Let $n$ be a positive integer.


Then:

$\ds \lim_{x \mathop \to 0^+} x^k \paren {\ln x}^n = 0$


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:

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

$\blacksquare$