Is $f(x)=\frac{x^{2}-1}{x-1}$ continuous at $x=1$? Given $f(x)=\frac{x^{2}-1}{x-1}$. The function is said to be discontinuous at $x=1$ but since we can simplify it and rewrite $f(x)=x+1$, this removes the discontinuity. So is the function continuous or discontinuous at $x=1$
How do the two forms of $f(x)$ differ as both expressions are equal to each other. What stops us from simplifying the earlier expression and saying the function is continuous ?There was a similar question here but it didn't address my latter point.

$\lim_{x\rightarrow 1} f(x) = 2$, so the singularity is removable.
Let $$g(x) = \left\{ \begin{array}{cc}  f(x), & x \ne 1\\ 2, & x=1 \end{array}  \right.$$  The function $g(x)$ is continuous.
So, yes, $f(x)$ is discontinous but this discontinuity is easily repaired.  If you were to graph $y=f(x)$, it would be the straight line $y=x+1$ with the point $(1,2)$ removed.
