What's the value of $\frac{(x-1)x^2}{x-1}$ when $x=1$?

Question

Consider the function $$f(x)= \frac{(x-1)x^2}{x-1},\quad x \in \Bbb R\ .$$ This function gives simply gives $f(x)=x^2$ by cancelling the term $x-1$, if I am not wrong. The variable here is $x$ which can take arbitrary value, we just cancel that term as for any $x$, suppose $x=0,2,3,4, \ldots,-1,-2,-3,\ldots$ then the term is going to get cancelled, but what about the value $1$? If we have $x=1$ there then the term would not be defined, so why do we always cancel the similar terms without thinking what values variable takes? I know this question is really very strange but I want to be clear about my approach. This question is somewhat which I never asked before.

Answer 1

The function is undefined at $x = 1$ because $\frac00$ is undefined. If you graphed this the graph would look like the graph of $g(x) = x^2$ with a point simply missing from the graph. We say that the limit of the function as $x$ tends to $x^2$ (=1) from either $x \gt 1$ or $x \lt 1$ tends to 1. ($\lim_{x \rightarrow 1} f(x) = x^2$)

We also refer to $x=1$ as a "removable singularity". If ever you factor and divide by $(x - 1)$ you must specify that you are presuming $x$ does not equal $1$ for the rest of your conclusion.

This is a very common occurrence and plays very heavily in calculus.

It's good that you caught it.

Answer 2

"without thinking what values variable takes..." - the answer to this is that you always should think about the values the variable(s) can take. A correct simplification is $$\frac{(x-1)x^2}{x-1}=x^2\ ,\quad\hbox{if $x\ne1$}.$$ If you have a textbook which does not make the point that $x\ne1$, my suggestion is - throw it out and buy a better textbook!