# 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
Gold Member
Looks right to me.

• Eclair_de_XII
Eclair_de_XII
Arr, thanks for the help and giving that counter-example that I overlooked.

Mentor
Define ##f:(-1,0)\rightarrow \mathbb{R}## by ##f(x)=-\ln(-x)##. Show that ##f## is unbounded.
It would have been simpler to work with ##g(x) = \ln(x)## on the interval (0, 1). It would have made the arithmetic a bit less tedious. This function is the reflection across both the x and y axes of the one you have. Both f and g are unbounded for x near 0.

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