# limits of natural logarithm

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

 $\displaystyle\ln{x}\;=\;\int_{1}^{x}\frac{1}{t}\,dt\qquad(x>0)$ (1)

and derives the

 $\ln{xy}\;=\;\ln{x}+\ln{y}$

which implies easily by induction that

 $\displaystyle\ln{a}^{n}\;=\;n\ln{a}.$ (2)

Basing on (1), we prove here the

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

 $\displaystyle\lim_{x\to+\infty}\ln{x}\;=\;+\infty\quad\mbox{and}\quad\lim_{x% \to 0+}\ln{x}\;=\;-\infty.$ (3)

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

 $\frac{d}{dx}\ln{x}\;=\;\frac{1}{x}\;>\;0$

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

 $\ln{x}>M\quad\forall x>2^{n}.$

Hence the first limit assertion is true. Now  $-M<0$.  If  $x>2^{n}$,  then  $\ln{x}>M$  and

 $0\;<\;\frac{1}{x}\;<\;2^{-n},\qquad\ln\frac{1}{x}\;=\;\int_{1}^{\frac{1}{x}}% \frac{dt}{t}\;=\;\int_{x}^{1}\frac{du}{u}\;=\;-\ln{x}\;<\;-M$

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

Title limits of natural logarithm LimitsOfNaturalLogarithm 2014-12-12 10:15:50 2014-12-12 10:15:50 pahio (2872) pahio (2872) 11 pahio (2872) Theorem msc 33B10 ImproperLimits GrowthOfExponentialFunction FundamentalTheoremOfCalculusClassicalVersion DifferentiableFunctionsAreContinuous