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.

Title	limits of natural logarithm
Canonical name	LimitsOfNaturalLogarithm
Date of creation	2014-12-12 10:15:50
Last modified on	2014-12-12 10:15:50
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Theorem
Classification	msc 33B10
Related topic	ImproperLimits
Related topic	GrowthOfExponentialFunction
Related topic	FundamentalTheoremOfCalculusClassicalVersion
Related topic	DifferentiableFunctionsAreContinuous