Is function $f$ continuous? $$f(x) = \begin{cases} x-1.5 & \text{ if } x<1\\\\ \dfrac{1-x}{x^2-1} & \text{ if } x>1\end{cases}$$
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?
 A: 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 $0$.
A: 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.
A: 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.  
