Is $\frac{x^2}{x}$ continuous?

Question

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.

Answer 1

It is not continuous at $0$ because it is not defined at $0$ but it can be extended by continuity at $0$.

Answer 2

$$ x \text{ continuous at $0$ } \iff \lim_{x\to 0^-}f(x)=\lim_{x\to 0^+}f(x)=f(0)$$

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

Answer 3

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 $x=0$.

Answer 4

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 $\mathbb R$.

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

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 $\hat{f}(0)=0$.