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 ?
 A: See what happens when you multiply a fraction on top and bottom by $0$:
$$\frac{5}{3} = \frac{0\cdot 5}{0\cdot 3} = \frac{0}{0}.$$
Ick.
When you convert your fraction from its original form to $\frac{x+1}{x-1}$, you're multiplying top and bottom by $x$.  This multiplication won't be valid if $x=0$, so you have to eliminate that value from the calculation.  
A: Your prior function is 
$$f(x)=\frac{1+\tfrac{1}{x}}{1-\tfrac{1}{x}}$$
and the first condition you must attend is $x\ne 0$ and after that you can simplify and get 
$$f(x)=\frac{x+1}{x-1}$$
and then you also must have $x\ne 1$.
A: Your rewrite of $f(x)=\frac{x+1}{x-1}$ is valid for $x\ne 0$; but at $x=0$ the original statement of the function is undefined, and so $0$ is not in the domain of the original function.
A: It's a good observation
The idea is that both the functions are not same. It looks that they are ( because you will say that you simplified it), but it is not the case . When you canceled $x$, actually you assumed $x$ to be non zero. So as the functions are not equal , there domain will certainly differ.
Hope I've made myself clear.
Thanks.
A: Ok, this is a prime example of rational functions. The first statement is correct. The domain of $f(x)$ is $\mathbb{R}-${0,1}. But as you have noted correctly the point at $x=0$ is just not defined in the function as stated in its initial form which is the definition of your function.
It is like saying what is the domain of $\frac{(x-1)^2}{x-1}$. How I defined it it is defined everywhere in $\mathbb{R}$, but the limit from right and left at $x=1$ is the same.
A: Note that if there is some thing like $f (x)=\ln (x^2) $  where $x\in R $ then we cant always write $f (x)=2\ln (x) $ as if the x is negative then the log isnt defined for negative reals. So we cant always manipulate a function to simplify it we always have to see the original form of the function. In your case if you multiply by x you are lerforming similar mistake like a very common example $0=0\implies 0.1=0.2\implies 1=2$ which is false. You cant multiply or divide by $0$.
A: 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
$$x\in D \implies f (x)=\frac {x+1}{x-1} $$
