What should be the solution for $\dfrac{(x-1)^2}{(x-1)} = 0$ where obviously that x should not be equal to 1? What should be the solution for $\dfrac{(x-1)^2}{(x-1)} = 0$, where it is obvious that x should not be equal to 1?
In this case, when can we only cancel the common factors on the numerator and denominator of a rational expression? Do we need to identify the numbers that are not possible after canceling the common factor?
This might be a simple question but often not properly discussed in most of the math class.
I am studying continuity of a function and I encounter this problem in my head while doing an analysis on determining if a certain function is continuous or not.
Please feel free to share your answers and I would like to thank you in advance for sharing your thoughts about this.
 A: There is no solution: $\lim_{x \to 1} \frac{(x-1)^2}{x-1} = 0$ which means the function gets arbitrarily close to $0$, but for no $x$ does $f(x) = 0$.
A: Strictly speaking $\frac{(x-1)^2}{x-1}$ has no solution.
Regarding "cancellation": It's not quite true that $\frac{(x-1)^2}{x-1}=x-1$. What's true is $$\frac{(x-1)^2}{x-1}=x-1\quad(x\ne1).$$
A: The given expression does not have a root, since the numerator vanishes exactly on the same points as the numerator. However we can extend it to the real line in order to make it continuous as follows:
\begin{align*}
f(x) =
\begin{cases}
x-1, & x\neq 1,\\
0, & x = 1
\end{cases}
\end{align*}
The thing is that we are dealing with a removable singularity.
Hopefully this helps!
A: $$ \frac{(x-1)^2}{x-1} = 0 $$
By cancellation it becomes,
$$ x-1=0 $$
However putting x as $1$ will give denominator $0$. Thus this expression has no solution
A: So, if
$$
f(x) = \frac{(x-1)^2}{(x-1)}
$$
then ${f(x)=0}$ does not have any solutions. The thing is, it's not true that ${f(x) = x-1}$; it's true that if ${x\neq 1}$, then ${f(x)=x-1}$, but we cannot perform the cancellation when ${x=1}$, it doesn't make any sense.
Instead, what we can do is "extend" ${f(x)}$ to make sense of it at $1$. It's what's called a "removable singularity". You may notice that
$$
f(x) = \frac{(x-1)^2}{(x-1)}
$$
behaves just like ${x-1}$ around the point ${x=1}$, and even satisfies ${\lim_{x\to 1}f(x) = 0 = (1-1)}$. And so it makes sense to extend ${f(x)}$ to the function ${x-1}$, which will now include ${x=1}$ as a valid part of the domain and it will remain a continuous function. Now you can say ${\tilde{f}(x) = 0}$ has solution ${x=1}$ (I have denoted ${\tilde{f}}$ here to mean "$f$ extended to include $1$"). Hopefully that helps.
