Domain of the function $f(x)=\frac{1+\tfrac{1}{x}}{1-\tfrac{1}{x}}$ The domain of the function $f(x)=\frac{1+\tfrac{1}{x}}{1-\tfrac{1}{x}}$ is said to be $\mathbb R-\{0,1\}$, given $f(x)$ is a real valued function. I understand why that is the case, since for both $1$ and $0$ the denimonator becomes $0$ and the value is undefined. But,
$$
f(x)=\frac{1+\tfrac{1}{x}}{1-\tfrac{1}{x}}=\frac{x+1}{x-1}
$$
Now I don't see any problem with $x$ taking the value $0$. What really is the domain of the function and how do I justify both the scenarios ?

The domain of $f $ is
$$D=\{x\in \mathbb R \;\;: \;x\ne 0 \land \frac {1}{x}\ne 1\}$$
but $$\frac {1}{x}=1\iff x=1$$
hence
$$D=\{x\in\mathbb R \;\;:\;x\ne 0\land x\ne 1\} $$
$$=\mathbb R \backslash \{0,1\} $$
$$=(-\infty,0)\cup (0,1)\cup (1,+\infty) $$
and
$$x\in D \implies f (x)=\frac {x+1}{x-1} $$
