<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] Let consider
$$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}$$
from here we need to set $x\neq1,-1,0$, morover we need that
$$\frac{2}{x+1} + \frac{1}{x}=\frac{3x+1}{x(x+1)}\neq 0\implies x\neq -\frac13$$
Simplifing the original expression we obtain
$$f(x)=\frac {\frac{1}{x-1} + \frac{1}{x+1}}{\frac{2}{x+1} + \frac{1}{x}}\to\bar f(x)=\frac{2x^2}{(3x+1)(x-1)}$$
which is well defined for $x\neq -\frac13,1$.
In conclusion the function $f(x)$ is not defined for  $x=1,-1,0,-\frac13$ and $x=0$ and $x=-1$ are two point with removable discontinuity.
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