limits of natural logarithm

The parent entry (http://planetmath.org/NaturalLogarithm) defines the natural logarithm as

$\ln x={\displaystyle {\int }_1^x}{\displaystyle \frac{1}{t}}dt\mathit{\hspace{1em}\hspace{1em}}(x>0)$(1)

and derives the

$$\ln xy=\ln x+\ln y$$

which implies easily by induction that

$\ln a^n=n\ln a.$(2)

Basing on (1), we prove here the

Theorem.  The function  $x\mapsto \ln x$ is strictly increasing and continuous on ${ℝ}_{+}$.  It has the limits

$\underset{x\to +\mathrm{\infty }}{lim}\ln x=+\mathrm{\infty }\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\underset{x\to 0+}{lim}\ln x=-\mathrm{\infty }.$(3)

Proof.  By the above definition, $\ln x$ is differentiable:

$$\frac{d}{dx}\ln x=\frac{1}{x}>\mathrm{\hspace{0.33em}0}$$

Accordingly, $\ln x$ is also continuous and strictly increasing.

Let $M$ be an arbitrary positive number.  We have  $\ln 2={\int }_1^2\frac{dt}{t}>0$.  There exists a positive integer $n$ such that  $n\ln 2>M$ (see Archimedean property).  By (2) we thus get  $\ln 2^n>M$, and since$\ln x$ is strictly increasing, we see that

$$\ln x>M\mathit{\hspace{1em}}\forall x>2^n.$$

Hence the first limit assertion is true.Now  .  If  $x>2^n$,  then  $\ln x>M$  and

(substitution (http://planetmath.org/SubstitutionForIntegration)  $xt:=u$).  From this we can infer the second limit assertion.