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Consider the function \begin{align*} f(x) = \frac{x^{3} - x^{2}}{x-1} \end{align*}

This function is not continuous at $x = 1$.

However, using factorization, this function is equal to $ \begin{align*} f(x) = \frac{x^2(x-1)}{x-1} = x^2 \end{align*}

which is continuous at $x = 1$!

So I am confused right now, is this function continuous or not?

Átila Correia
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  • The expression $(x^3-x^2)/(x-1)$ is not defined when x=1, so you need to separately define what the value of the function is for x = 1. If f(1) = 1 then the function is continuous. – MilesB Oct 02 '21 at 20:40

1 Answers1

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What is the definition of continuity in $\mathbb{R}$?

We say that a function $f:X\to\mathbb{R}$ is continuous at $a\in X$ iff \begin{align*} (\forall\varepsilon > 0)(\exists\delta_{\varepsilon}>0)(\forall x\in X)(|x - a| < \delta_{\varepsilon} \Rightarrow |f(x) - f(a)| < \varepsilon) \end{align*}

At the given example, we have the function $f:\mathbb{R}\backslash\{1\}\to\mathbb{R}$, which is not defined at $x = 1$.

Hence it cannot be continuous at $x = 1$.

On the other hand, we have the following definition of limit in $\mathbb{R}$.

A function $f:X\to\mathbb{R}$ has limit at $a$, where $a$ is an accumulation point of $X$, and it equals $L\in\mathbb{R}$ iff \begin{align*} (\forall\varepsilon>0)(\exists\delta_{\varepsilon} > 0)(\forall x\in X)(0 < |x - a| < \delta_{\varepsilon} \Rightarrow |f(x) - L| < \varepsilon) \end{align*}

As you can see, in the limit definition we do not require that $a$ is a point of $X$.

Precisely, we are interested in the behaviour of $f$ arbitrarily close to $a$, but not necessarily equal to $a$.

Therefore, as you have noticed, the proposed limit exists even though $f$ is not continuous.

That is because $1$ is an accumulation point of $f$, thus we can divide by $x - 1$.

Moreover, as @MilesB has noticed in the comments, $x = 1$ is a removable singularity.

So, if you define $f(1) = 1$, it becomes continuous.

Átila Correia
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  • In logic, $f(x) = (x^3 - x^2)/(x -1)$ is just a formula describing a property of $f$ and tells you nothing about the domain of definition of $f$. There seems to be a widespread view in the teaching of elementary mathematics that the formula should be read as implying $f$ has the largest domain that makes the right-hand side well-defined. I think this is a bad trend. On the one hand, you should be explicit about the intended domains of functions, while on the other $(x^3-x^2)/(x-1)$ and $x^2$ represent the same rational function of $x$ (in $\Bbb{R}(x)$). – Rob Arthan Oct 02 '21 at 21:23
  • Thanks for the contribution. I have indeed assumed that the domain of $f$ is the largest possible set which makes its expression valid. So the domains should be $D_{f} = \mathbb{R}\backslash{1}$ and $\mathbb{R}$ (after removing the singularity). – Átila Correia Oct 02 '21 at 21:26
  • Sure (and +1, by the way): I wasn't criticising your answer but rather the person who set the problem that the OP is trying to solve. You've given the right answer under the common reading. I just think we should be more careful when defining functions. – Rob Arthan Oct 02 '21 at 21:33
  • Thanks for the upvote! I did not mean to be impolite. I appreciate your observation. – Átila Correia Oct 02 '21 at 21:36
  • I didn't think you were impolite to me at all, but rather the other way round $\ddot{\smile}$, as I didn't make it clear what I was criticising. – Rob Arthan Oct 02 '21 at 21:54