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If you have a function $f(x)=\dfrac{x^2}{x}$, then is the function continuous at $x=0$?

On one hand, if you simplify it and end up with $f(x)=x$, it is continuous at $0$, but if you keep it in its original form, at $x=0$ the function is not defined.

AlexR
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user249990
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    You already said it. Since it's not even defined at $0$, it can't be continuous at $0$. It can however be continuously extended to $0$ (by defining $f(0) := 0$) – AlexR Jun 23 '15 at 09:19
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3 Answers3

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By definition, a function is continuous at a point $a$ if $f(a)$ is defined and $$\lim_{x\to a}f(x)=f(a).$$ As you already noted, your function is not defined at $x=0$, and is therefore not continuous.

It is however possible, as pointed out in a comment, to extend the function to a continuous one by simply defining $f(0)=0$.

Mårten W
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    It's worth adding that, in some contexts, one assumes that any possible extensions of this type have been made, irrespective of the literal interpretation of function is written: For example, often (and perhaps objectionably) that $x \mapsto \frac{\sin x}{x}$ is often used to denote the function $f$ given by that rule at $x \neq 0$ and is defined to be $1$ at $x = 0$. This is especially common when working with power series representations of such functions (e.g., of $x \mapsto \frac{e^x - 1}{x}$) and complex analysis; in the latter setting such 'holes' are called removable singularities. – Travis Willse Jun 23 '15 at 09:33
  • @Travis: That is a good point. In practice, we are sometimes not very strict about removable singularities and the domain of the particular function, but there are situations when details such as these matter. – Mårten W Jun 23 '15 at 09:43
  • Yes, agreed, and those situations definitely include when students are learning the relevant definitions, as looks like the case here. (+1 for the nicely put answer, btw.) – Travis Willse Jun 23 '15 at 09:49
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If you talk about the expression $\frac{x^2}{x}$ then you are right in saying that it cannot be evaluated in that form at 0. In such cases when an algebraic expression of a function is not applicable you look for extensions that are either continuous, $\mathcal{C}^1$ or $\mathcal{C}^2$ or whatever property you want to preserve.

In this case you merely cancel out the $x$ from the denominator and the expression $f(x)=\frac{x^2}{x}$ is equivalent to $f(x)=x$. Hence $f(x)=x$ is an extension of $f(x)=\frac{x^2}{x}$ to 0.

Miz
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It is not by the definition of continuity. $\lim \limits_{x \to 0} \frac{x^2}{x}$ is not defined at the absolute value of $x=0$ even though its $RHL$ and $LHL$ do exist at that point both being equal to $0$. It's a kind of removable continuity where by certain adjustments the function can be made continuous.

Suraj S
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