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$$f(x) = \begin{cases} x-1.5 & \text{ if } x<1\\\\ \dfrac{1-x}{x^2-1} & \text{ if } x>1\end{cases}$$

Is this function continuous or not? Some people say it is because the value $x=1$ is not defined for the function while others say it is not continuous because it has a hole at $x=1$. Which is correct?

Rodrigo de Azevedo
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  • See https://math.stackexchange.com/questions/1482787/can-we-talk-about-the-continuity-discontinuity-of-a-function-at-a-point-which-is/1482900#1482900 – Michael Hoppe Sep 13 '17 at 09:13
  • A function is continuous at a point $a$ if its value $f(x)$ approaches the value $f(a)$. Thus, if it is not defined at $a$, it cannot be continuous at that point. – Mauro ALLEGRANZA Sep 13 '17 at 09:16
  • A function $f : E → \mathbb R$ is continuous on the set $E$ if it is continuous at each point of $E$. – Mauro ALLEGRANZA Sep 13 '17 at 09:21

3 Answers3

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It is continuous, but that means only that it's continuous on it's domain, that is for all $x\ne 1$

It is not continuous outside it's domain, that is it's not continuous for all $\mathbb R$ - it's because in order to be continuous somewhere it has to be actually defined there.

Note that failing to be defined is an important exception as it will invalidate the prerequisite of the intermediate value theorem. For example $1/x$ is continuous on it's domain, but it's not continous for all $[-1,1]$ which means that it's not required by the theorem to take the intermediate value $0$.

skyking
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2

The function, as it stands now with definitions for $x<1$ and $x>1$ and neither $x\leq1$ nor $x\geq1$, is undefined at $x=1$. Thus it is neither continuous nor discontinuous for $x=1$; it simply isn't. At all other points, it is continuous.

It is also worth noting that if you fill in the gap with $f(1)=-0.5$, then the function becomes defined and continuous on the whole number line.

Arthur
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It could be continuous at $x=1$ if the function was defined for $x$ is greater than and equal to 1, or less that or equal to 1, because from the right, $$f(x)=\frac{1-x}{x^2-1}=\frac{1-x}{(x-1)(x+1)}=\frac{-1}{1+x}$$

At this value, from the right, as $x$ tends to 1, is $-0.5$

From the left, $f(x)=1-1.5=-0.5$

Thus, the limit from the left equals the limit from the right.

Harry Alli
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    I guess you mean $f$ can be defined in such a way at $x=1$ that $f$ becomes continuous at $x=1$. Because at it's present form, $f$ is not even defined at $x=1$. – Krish Sep 13 '17 at 09:13
  • It's not continuous at $x=1$ because it's not even defined there. – skyking Sep 13 '17 at 09:14
  • Indeed, will edit it. Sorry, I thought he had made one of the bounds inclusive. – Harry Alli Sep 13 '17 at 09:14