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

• Eclair_de_XII
In summary: You can see this by plotting them on the same graph and seeing that as x goes to 0, both f and g go to infinity.
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?

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

vela and Delta2
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:
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*}

Is r inside the interval you are supposed to be working on?

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:
Looks right to me.

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

Eclair_de_XII said:
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.

FactChecker

## 1. What is ln() and why is it unbounded near zero?

ln() is the natural logarithm function, which is the inverse of the exponential function. It is unbounded near zero because as the input value approaches zero, the output value approaches negative infinity.

## 2. What does it mean for a function to be unbounded?

A function is unbounded if its output values have no upper or lower limit. This means that as the input values increase or decrease, the output values also increase or decrease without bound.

## 3. Why is it important to understand why ln() is unbounded near zero?

Understanding why ln() is unbounded near zero is important in various fields of science and mathematics, such as calculus, statistics, and physics. It allows us to accurately model and analyze natural phenomena that involve exponential growth or decay.

## 4. Can you provide an example of a real-life situation where ln() being unbounded near zero is relevant?

One example is radioactive decay, where the amount of a radioactive substance decreases exponentially over time. The rate of decay can be described using ln(), and understanding its behavior near zero is crucial in predicting the amount of time it takes for the substance to decay completely.

## 5. Is there a way to work around the issue of ln() being unbounded near zero?

Yes, there are ways to work around this issue, such as using a different base for the logarithm function or using a different function altogether. However, it is important to understand and acknowledge the behavior of ln() near zero in order to make accurate calculations and interpretations in scientific and mathematical contexts.

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