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

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

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Answer 1 · 2015-10-22T15:45:26

Since $f (x) = \frac{(x - 1) (x + 1)}{x^{2} - 1}$ $f(x) = ((x-1)(x+1))/(x^2-1)$ 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 $x = \pm 1$ $x=+-1$

Explanation:

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

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

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

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