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?

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.  
