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.
 A: No.  No.  No.
$f$ is not merely discontinuous at $x=1$.  It is undefined and doesn't exist and $x=1$ is not in the domain.  $f(x)$ and $x=1$ does not exist.
"since we can simplify it and rewrite f(x)=x+1"
We can do no such thing at all.
$g(x) = x+1$ is an completely different thing than $f(x) = \frac {x^2-1}{x-1}$.  The difference is $x+1$ exists at $x=1$.  But $f(x)=\frac {x^2-1}{x-1}$ does not exist at $x=1$.  As they do different things at $x=1$ they are not the same thing at all.  That's all.  If two functions have different domains they are different things.  That they take the same values on the points of their domains in common is irrelevant.  They have different domains. End of story.
"How do the two forms of f(x) differ as both expressions are equal to each other."
$f(x)$ does not have two forms and the expressions $\frac {x^2-1}{x-1}$ and $x+1$ are not equal to each other.  At $x=1$ we have $x+1=2$.  But at $x=1$ we have $\frac {x^2-1}{x-1}$ is non-existent.  It DOES NOT EXIST!
"What stops us from simplifying the earlier expression and saying the function is continuous ?"
The same thing that stops us from switching bean sprouts with chocolate syrup and saying the snack is healthy.  Just because the snacks are exactly the same at $ingredient \ne bean\ sprouts$, the are different at $ingredient = bean\ sprout$ so the snacks are different things.
$x+1$ and $\frac {x^2-1}{x-1}$ are different things because one is defined at $x+1$ and the other isn't.  As they are different things they are .... different things.
If a guy says "I'm not a doctor but I play one on TV, now let's look at that appendix"... run...
A: $\lim_{x\rightarrow 1} f(x) = 2$, 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.
A: You can only simplify the expression assuming that $x\neq 1$. Since on $x=1$, the function is undefined, therefore the function is discontinuous at $x=1$.
