What is the domain of $f(x)=\frac{1}{x}-\frac{1}{x}$? Question: What is the domain of $f(x)=\frac{1}{x}-\frac{1}{x}$? Does the function have a removable discontinuity at $x=0$?
My attempt: My first intuition told me that it was $\mathbb R$, since we just have $f(x)=0$. However, when $x=0$, we get $f(x)= \mathrm{undefined} - \mathrm{undefined}=\mathrm{undefined}$. 
So I think that it is rather $\mathbb R_{\neq0}$. If that is correct, can we call $x=0$ a removable discontinuity?
 A: Well as you mentioned you get the domain of $f(x)$ as $\Bbb{R}-\{0\}$ and the range for the given function will just be zero.
So the domain will be

The range will be

A: Go for domain before simplifying .
  for example : $y=f(x) :\mathbb{R}\rightarrow \mathbb{R}$
$$y=(\sqrt{x})^2\\ $$if you simplify $y=x , \mathrm{domain}=\mathbb{R}$ 
but it's wrong $$\mathrm{domain}=\left \{ x|x \geq 0 \right \}$$ because of $\sqrt{x}$.
A: The domain of a function needs to be specified as part of the definition of the function. A formula for a function does not define the domain of the function, unless we specify the domain in some way from the formula: for example, we might choose the specification
"the domain of $f$ is the set of real numbers $x$ such that the given formula for $f(x)$ is well defined".
But this is not a definition of "the domain of $f$"; it is just one of many possibilities for the domain. Thus your question is well defined only if we add to it some specification of the domain. If we chose the specification above, then the answer is $\Bbb R\setminus\{0\}$ or, in your excellent notation, $\Bbb R_{\neq0}$. Zero is indeed a removable discontinuity in this case.
