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

$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$
$=\frac{\frac{2x}{(x-1)(x+1)}}{\frac{3x+1}{x(x+1)}}$
$=\frac{2x^2(x+1)}{(x-1)(x+1)(3x+1)}$
$=\frac{2x^2}{(x-1)(3x+1)}$
so $f(x)$ can be defined except at $x=1$ and $x=-\frac{1}{3}$. The apparent singularities at $x=0$ and $x=-1$ are removable.
