Order of Natural Logarithm Function

Theorem

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

Then:

$\ln x = \map \OO {x^\epsilon}$ as $x \to \infty$

and:

$\ln x = \map \OO {x^{-\epsilon} }$ as $x \to 0^+$

where $\OO$ is big-O notation.

Proof

We first show that:

$\ln x = \map \OO {x^\epsilon}$ as $x \to \infty$

We show that for $x \ge 1$, we have:

$\ds 0 \le \ln x \le \frac 1 \epsilon x^\epsilon$

We first show that for $t \ge 0$, we have:

$\ds t \le \frac 1 \epsilon e^{\epsilon t}$

The claim then follows taking $t = \ln x$.

Define a real function $f : \openint 0 \infty \to \R$ by:

$\ds \map f t = \frac 1 \epsilon e^{\epsilon t} - t$

for $t \ge 0$.

Then $f$ is differentiable with derivative:

$\map {f'} t = e^{\epsilon t} - 1$

by Derivative of Exponential Function.

So:

$\map {f'} t \ge 0$ for all $t \ge 0$.

So, from Real Function with Positive Derivative is Increasing:

$f$ is increasing for $t \ge 0$.

That is, for $t \ge 0$:

$\ds \map f t \ge \map f 0 = \frac 1 \epsilon > 0$

So:

$\ds \frac 1 \epsilon e^{\epsilon t} - t > 0$

for $t \ge 0$.

So:

$\ds \frac 1 \epsilon e^{\epsilon t} \ge t \ge 0$

for $t \ge 0$ as required.

Hence, for $x \ge 1$ we have:

$\ds 0 \le \ln x \le \frac 1 \epsilon x^\epsilon$

From the definition big-O notation, we have:

$\ln x = \map \OO {x^\epsilon}$ as $x \to \infty$

We now show that:

$\ln x = \map \OO {x^{-\epsilon} }$ as $x \to 0^+$

In particular, we can show that:

$\ds \size {\ln x} \le \frac 1 \epsilon x^{-\epsilon}$

for $0 < x \le 1$.

Note that since:

$\ds \ln x \le \frac 1 \epsilon x^\epsilon$

for $x \ge 1$ we have:

$\ds \map \ln {\frac 1 x} \le \frac 1 \epsilon x^{-\epsilon}$

for $0 < x \le 1$.

So, by Logarithm of Power:

$\ds -\ln x \le \frac 1 \epsilon x^{-\varepsilon}$

For $0 < x \le 1$, we have:

$\ln x \le 0$

so:

$\size {\ln x} = -\ln x$

from the definition of the absolute value.

So we obtain:

$\ds \size {\ln x} \le \frac 1 \epsilon x^{-\epsilon}$

for $0 < x \le 1$ as required.

From the definition big-O notation, we therefore have:

$\ln x = \map \OO {x^{-\epsilon} }$ as $x \to 0^+$

$\blacksquare$