Is this a valid explanation of why ln() is unbounded near zero?

Eclair_de_XII
Homework Statement:
Define ##f:(-1,0)\rightarrow \mathbb{R}## by ##f(x)=-\ln(-x)##. Show that ##f## is unbounded.
Relevant Equations:
A function ##f## is said to be unbounded if for all positive numbers ##M##, there is a ##y## in ##\textrm{dom}(f)## such that ##|f(y)|>M##.
So far, I found the derivative of ##f##:

\begin{align*}
\frac{d}{dx}\,f(x)&=&-\frac{d}{dx}\,\ln(-x)\\
&=&-\left(\frac{1}{(-x)}\right)(-1)\\
&=&-\frac{1}{x}
\end{align*}

##f'(x)## is always positive and never zero on its domain.

Hence, ##f## does not have a local maximum and is always increasing on the interval ##(-1,0)##.

Are these conditions sufficient to argue that ##\ln## is unbounded near zero?

Staff Emeritus
Gold Member
What about f(x)=x? The derivative is always positive as well. Is that unbounded?

vela and Delta2
Homework Helper
Gold Member
That is not what they are looking for. You should directly use the definition of "unbounded" that you gave. Assume that you have a value for ##M \gt 0## and determine a region ##R=(0,r]## for which ##x\in R## implies ##ln(x)\gt M##.

EDIT: The above is wrong. Determine a value, ##r \in (-1,0)## where ##|f(r)|\gt M##.

Last edited:
Eclair_de_XII
It's not as simple as setting ##f(r) = M + 1## and then solving for ##r##, is it?

##r=-\exp[-(M+1)]##

\begin{align*}
f(r)&=&-\ln(-r)\\
&=&-\ln[-(-\exp[-(M+1)])]\\
&=&-\ln[\exp[-(M+1)]]\\
&=&\ln[\exp(M+1)]\\
&=&M+1\\
&>&M
\end{align*}

Staff Emeritus
Gold Member
Is r inside the interval you are supposed to be working on?

Eclair_de_XII
Let me see:

1. The ##\exp## function is always positive, so ##-\exp(x)## for some ##x## is always negative.
2. ##e>2>0##. ##M## is positive, so the sum ##M+1## is positive. ##|e^{-(M+1)}|=\frac{1}{e^{M+1}}<1##, as a result.

I should think that ##r=-\exp[-(M+1)]## is in ##(-1,0)##

Last edited:
Staff Emeritus