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This might seem as a weird and straight forward to answer question, but it really confuses me. As the title states, I'm wondering whether

$$f(x)=\frac{x^2}{x}$$

is continuous. To be more precise, I'm wondering whether it's continuous at $x=0$. That question arose because I kind of have two contradicting ideas regarding that situation.

Contiuity definition

In university, one of the definitions we got was that a function is continuous in $x_0$ (here $0$) if (and only if) the following is true: $$\lim\limits_{x\searrow x_0}f(x)=\lim\limits_{x\nearrow x_0}f(x)=f(x_0)$$ The first two parts are obvious - of course they both approach $0$. But the last part is what's causing me trouble here - because it would yield to $\frac{0^2}{0}=\frac{0}{0}$ which is known not to be defined. So $f(x)$ wouldn't be continuous in $x_0=0$, right?

Arithmetic

The obvious thing to do though would be to rewrite $f(x)$ as $$\widetilde f(x)=x$$ Now it'd be perfectly clear that $f(0)=0$ just as well. So is the function continuous or is it not?

My approach

So I have an idea of what's probably happening here, but I'm really not sure about it, so I'd be happy for your help. In my understanding, $f(x)$ is defined as $f:\mathbb{R}\setminus\{0\}\rightarrow\mathbb R,\ x\mapsto\frac{x^2}{x}$, whereas $\widetilde f(x)$ is defined as $\widetilde f:\mathbb R\rightarrow\mathbb R,\ x\mapsto x$. So both functions would be continuous, but on another domain. However, that's still weird, isn't it? I mean, didn't we only make equivalence transformations, meaning that all parts of the following would be true? $$f(x)=\widetilde f(x)\qquad\Longleftrightarrow\qquad \frac{x^2}{x}=x$$

$\varepsilon-\delta$ criteria

And then there's another criteria of continuity, called the $\varepsilon-\delta$ criteria. I'm not sure if I can explain it well in english, so I'll just write it down in predicate logic: $$\forall\varepsilon>0\exists\delta>0\forall x\in D,|x-x_0|<\delta:|f(x)-f(x_0)|<\varepsilon$$ So this would once again apply for $\widetilde f(x)$ and not for $f(x)$. But then again we learned a visualisation of this criteria, with a rectangle of dimensions $2\delta\times2\varepsilon$ closing in around $x_0$ - this would work quite fine with both variations.

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  • $\begingroup$ The answer is straightforward: it's not continuous at $0$ because $0$ is not inside the domain of $f$. That's the end of it. I remember that this confused me too for a while. $\endgroup$ – rubik Jan 26 at 22:13
  • $\begingroup$ @rubik ok, but isn't dividing by $x$ an equivalent transformation? And if so, does that imply that equivalent transformations can alter the domain? $\endgroup$ – MetaColon Jan 26 at 22:17
  • $\begingroup$ It is not because you cannot divide by $x = 0$. $\endgroup$ – Klaus Jan 26 at 22:18
  • $\begingroup$ @Klaus ok than I now also know where my mistake lay. $\endgroup$ – MetaColon Jan 26 at 22:19
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It is not continuous at $0$ because it is not defined at $0$ but it can be extended by continuity at $0$.

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  • $\begingroup$ To be quite honest, this doesn't help me at all $\endgroup$ – MetaColon Jan 26 at 22:09
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    $\begingroup$ But this is the right answer. For a function to be continuous at point a, the value of f(a) must be the limit of f(x) when x goes to a. In your case there is no f(0), so it cannot be continuous there. If you define f(0)=0 then you are good $\endgroup$ – Shaq Jan 26 at 22:11
  • $\begingroup$ @Shaq but imagine you don't define that. Isn't it still exactly the same as $\widetilde f$? $\endgroup$ – MetaColon Jan 26 at 22:12
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    $\begingroup$ $f$ and $\tilde{f}$ are not the same functions because the domains are different! $\endgroup$ – Klaus Jan 26 at 22:14
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    $\begingroup$ @MetaColon No, it's not. The original function cannot be defined at zero as it is, whereas $\;\tilde f\;$ is defined everywhere. This is the right answer. $\endgroup$ – DonAntonio Jan 26 at 22:14
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Actually, if we are going to be rigorous, then the question makes no sense. That's so because you did not state what is the domain of $f$. If you choose to decide that the domain is the largest subset of $\mathbb R$ for which the expression $\dfrac{x^2}x$ makes sense (which is $\mathbb{R}\setminus\{0\}$), then the answer now becomes: it makes no sense to ask whether $f$ is continuous at $0$ since continuity is defined only for points of the domain.

However, if the domain of $f$ is $\mathbb{R}\setminus\{0\}$, then you can extend $f$ to a continuous function from $\mathbb R$ into $\mathbb R$.

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  • $\begingroup$ I know you meant this, but I want to add: if the domain is $\Bbb R \setminus 0$, then for this particular $f$, there is a continuous extension to $\Bbb R$. In general, not all functions on the first set admit continuous extension to the second. Example: $g(x) = \frac{1}{x}$. $\endgroup$ – John Hughes Jan 26 at 22:23
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$$ x \text{ continuous at $0$ } \iff \lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)=f(0)$$

But $f(0)$ doesn't exist, so this doesn't hold.

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The definition of continuity ends with $\cdots = f(x_0).$

Since $f(0)$ doesn't exist, the function is not continuous there. The graph of the function is a straight line of slope $1$ through the origin, but has a hole at the origin.

To be continuous at a point, there are three things necessary:

  1. The limit exists.
  2. The function exists.
  3. The above two are equal.
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You are asking if the function is continuous at $x=0$. The definition of being "continuous at $x=0$" means that $$\lim_{t\to0}f(t)=f(0)$$ With this function, $\lim_{t\to0}f(t)=0$ and $f(0)$ is undefined. So $$\lim_{t\to0}f(t)\neq f(0)$$ and therefore the function is not continuous at $x=0$.

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When you give a definition of a function, it is down to you to make clear what the intended domain of that function is. There is a common convention that if you write:

$$ f(x) = t(x) $$

where $t(x)$ is some algebraic expression in the variable $x$, then the domain of $f$ is restricted to values of $x$ for which deriving the value of $t(x)$ is well-defined, e.g., doesn't involve division by zero.

This convention is very vague and in itself ill-defined. E.g., the domain of the function $r$ defined by $$ r(x) = \sqrt{x} $$ will be at most $\Bbb{R}_{\ge0}$ if the range of $r$ is expected to be $\Bbb{R}$, but could be $\Bbb{C}$ if the range of $r$ is expected to be $\Bbb{C}$.

So when you ask about whether $f(x) = \frac{x^2}{x}$ is continuous, you are:

  • firstly, making a category error (an equation is not a function) and,
  • secondly, if we forget the category error and assume you mean the function $f$ defined by the equation, not giving enough information about the function in question.

A rigorous definition of your function would say precisely what its domain is intended to be.

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    $\begingroup$ What exactly do you mean by category error? $\endgroup$ – MetaColon Jan 26 at 22:43
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    $\begingroup$ $f(x) = x^2/x$ is an equation in which $f$ appears as a function. Your question is asking if the equation is continuous, but it's actually the function that may or may not be continuous. A category error is applying a notion (like continuity) to the wrong kind of thing (like an equation rather than a function). $\endgroup$ – Rob Arthan Jan 26 at 22:50
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The function is undefined at $x=0$ so can't be continuous there.

The definition of the function can be extended so that the function is given a value at $x=0$ in a way which does make the extended function continuous there.

If the function were extended by giving a value at $x=0$ which does not make it continuous, we would say that it had a "removable discontinuity" at $x=0$. The discontinuity would be removed by giving it "the correct value" at an isolated point.

Sometimes this language is used more informally to say "the definition can be repaired or extended" in the case where the function is not properly defined at a some isolated point.

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The question is not well-posed. It only makes sense to consider continuity at points in the domain of the function. If we view the function as $f\colon \mathbb{R}\setminus \{0\}\to \mathbb{R}$, then the function is continuous everywhere. This function has an extension $\hat{f}\colon \mathbb{R}\to \mathbb{R}$ which is continuous at zero. Simply define $\hat{f}(0)=0$.

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