<s>[INST] Why is $f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$ defined at -1, 0 and not at 1 Why does the function:
$$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$$
Is defined at $x = -1, x = 0$ (we will have $1/0$ in the fraction then) and not at $x = 1$ according to the following graph?

Live example
 [/INST] $f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$
$f(x)=\frac{(x-1)(x+1)(x)}{(x-1)(x+1)(x)} \cdot \frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}} $
$f(x)=\frac{(x+1)(x)+(x-1)(x)}{2(x-1)(x)+(x-1)(x+1)}$
$f(x)=\frac{x^2+x+x^2-x}{2x^2-2x+x^2-1}$
$f(x)=\frac{2x^2}{3x^2-2x-1}$
$f(x)=\frac{2x^2}{(3x+1)(x-1)}$
So the vertical asymptotes at $x=-\frac{1}{3}$ and $x=1$
while there only holes at $x=0$ and $x=-1$
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