Why does $\sqrt{\frac{x^2-2x+1}{x^2-1}}$ exist at $x = 1$?

Question

So there is this question of the finding domain of the following function:

$\sqrt{\frac{x^2-2x+1}{x^2-1}}$

Through graphing technology, it states that this function is existing at $x = 1$. However, wouldn't that be not possible because of the division by zero in the numerator $x^2-1$? Does the square root affect it in any way? Thanks.

Answer 1

For $x>1$, $f(x)=\sqrt{\frac{x-1}{x+1}}$. Strictly speaking, $f(x)$ is not defined at $1$, but we commonly identify functions with their "maximal continuous extension" like this. In other words we "fill in the holes" that may be present due to domain constraints.

Answer 2

Because $\sqrt{\frac{x^2-2x+1}{x^2-1}}=\sqrt{\frac{(x-1)^2}{(x-1)(x+1)}}=\sqrt{\frac{(x-1)}{(x+1)}}$ and the later is existing at $x=1$