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