Does $f(x)=((x-1)(x+1))/(x^2-1)$ have an asymptote or a hole?

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Answer 1 · 2015-10-22T15:45:26

Since $f(x) = ((x-1)(x+1))/(x^2-1)$ is not a curve, it does not have an asymptote (using most common definitions of the term.

It is undefined (has [in this case "removable"] holes) for $x=+-1$

Except for the values $x=1$ and $x=-1$ where $f(x)$ becomes equivalent to $0/0$

$f(x)=1$ (since $(x-1)(x+1) = (x^2-1)$ )

Does f(x)=((x-1)(x+1))/(x^2-1)f(x)=((x-1)(x+1))/(x^2-1) have an asymptote or a hole?