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I was discussing with this person online because they (basically) claimed that $$f(x)=\frac{x}{x^x}$$ is equal to 0 at $x=0$.

I argued this was not technically true because in order to get to that result you would have to "simplify" the function the following way:

$$ \begin{align} f(x)&=\frac{x}{x^x}\\ &=\frac{1}{x^{x-1}}\\ \end{align} $$

And so, when evaluating at $x=0$ we get $f(0)=0$.

We both agree to this point, but I'm arguing that that's taking the limit as $x\rightarrow 0^+$. He, on the other hand, argues that this is, in fact, not taking the limit but the value of the function itself.

We are now curious and don't really want a short answer. Thanks in advance!

Shotem
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6 Answers6

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In order to discuss continuity and discontinuity of a function $f$, it has to be defined there. If $f(x)=\dfrac{x}{x^x}$, then $f(0)=\dfrac{0}{0^0}$. Whether or not this expression has any meaning depends on whether you leave $0^0$ undefined, or define it as $1$:

  • If you leave $0^0$ undefined, then $f$ is undefined at $0$, and so it whether it is continuous at $0$ is also undefined. To understand why, consider that the usual definition of continuity of $f$ at $a$ is $$ \DeclareMathOperator{\epsilon}{\varepsilon} \DeclareMathOperator{\dom}{dom} \forall\epsilon>0:\exists\delta>0:\forall x\in\dom(f):|x-a|<\delta\implies|f(x)-f(a)|<\epsilon \, . $$ For this definition to make any sense, we need $f(a)$ to be defined. However, we can define a new function $\hat{f}$ that is continuous at $0$: $$ \hat{f}(x)= \begin{cases} \dfrac{x}{x^x} &\text{if $x>0 \, $,} \\[4pt] 0 &\text{if $x=0 \, $.} \end{cases} $$
  • If you define $0^0$ as $1$, then $f(0)=0$, and you can discuss continuity of $f$ at $0$. Since $$ \lim_{x \to 0^+}\frac{x}{x^x}=\frac{\lim_{x \to 0^+}x}{\lim_{x \to 0^+}x^x}=\frac{0}{1}=0 \, , $$ $f$ is continuous at $0$.

In any case, the simplification $f(x)=\dfrac{x}{x^x}=\dfrac{1}{x^{x-1}}$ is only valid when $x>0$, as $0^{-1}$ is always left undefined.

Joe
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The formula $f(x) = x/x^x$ is undefined at $x=0,$ since if you plug in $f(0) = 0/0^0$ you get nonsense. However, you can define $f(0)$ to be whatever you'd like. A function is not a formula, it is merely a collection of input output pairs.

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    I disagree with this answer. Why should $\frac{0}{0^0}$ be nonsense? $0^0$ is very often defined as $1$, and with good reason. – Joe Jul 04 '21 at 17:39
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    @Joe: But very seldom is this done in Calculus, as it runs the risk of introducing multiple exceptions to limit laws. – Arturo Magidin Jul 04 '21 at 17:46
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    @ArturoMagidin It is done all the time in calculus with series. Like how the series expansion for $e^x$ is $\sum_{n=0}^{\infty}\frac{1}{n!}x^n$, even at $x=0$ where that has a $0^0$ in it. – 2'5 9'2 Jul 04 '21 at 17:54
  • @alex.jordan: I teach calculus all the time; that's not my experience. This is, at best, an arguable position, but hardly a settled fact as Joe (and you?) seem to take it. – Arturo Magidin Jul 04 '21 at 17:56
  • @ArturoMagidin: I'm not saying that $0^0$ is always defined as $1$ in all contexts. But calling it nonsense is a step too far. – Joe Jul 04 '21 at 18:00
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    Sorry, I was just disputing the "very seldom done in calculus" part. Do you write Taylor series like $1+\sum_{n=1}c_nx^n$ to avoid having $0^0$ arise? That would be reasonable. – 2'5 9'2 Jul 04 '21 at 18:04
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The function's formula is $\frac{x}{x^x}$, and if you want to investigate its propoerties at $x=0$, you should work with that originally given formula for the function. Simplification to another formula is often only valid on a subset of the domain.

So $f(0)=\frac{0}{0^0}$. You have to confront what you want $0^0$ to mean. There are several valid positions to take regarding the value of $0^0$. See a post on this site about it.

One position is that $0^0$ is an empty product, and therefore equal to $1$, regardless of the base of that expression being $0$. In that case $f(0)=\frac{0}{1}=0$. If this is the position your friend takes, it is correct to conclude $0$ is the actual value of $f(0)$.

But of course $0^0$ might mean something else to someone else. And depending on what it means to them, then $f(0)$ is probably undefined.

2'5 9'2
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If we write your functions as such: $$f(x) = x\,e^{-x\log x},$$ then we see that the value of your function at $x=0$ is undefined because $\log 0$ does not exist. However, if we take the limit as $x\to 0^+$ of our function, we find that $\lim_{x\to 0^+}f(x)=0$. You can now define $f(0)=0$, and your function will be right-continuous at $x=0$. Is it not continuous there because the left limit does not exist since your function isn't defined for negative $x$.

Dispersion
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$$L=\lim_{x\to0}f(x)$$ $$f(x)=\frac{x}{x^x}=x^{1-x}=e^{(1-x)\ln(x)}$$ $$g(x)=\ln f(x)=(1-x)\ln(x)$$ $$\ln(L)=\lim_{x\to0}g(x)=-\infty$$ so: $$L=\lim_{y\to-\infty}\exp(y)=0$$ however, this is the limit from above

Henry Lee
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Re-write as $\frac{x}{\exp(x\log(x))}$.

The function $\exp$ is everywhere continuous. By L’Hopital’s rule, $x\log(x)$ tends to 0. Therefore the denominator of the original fraction tends to 1. Since the numerator tends to 0, the fraction tends to 0.

Setting the value of f(0) to 0 makes f continuous for x positive (well, left continuous at 0).

colin
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