Domain of $\frac{1}{x^{-1}}$

Question

Why WolframAlpha and other online calculators compute $x\in \mathbb{R}-\{0\}$ (and not $\mathbb{R}$) as the domain of $\frac{1}{x^{-1}}$?

Since $\frac{1}{x^{-1}}=\frac{1}{\frac{1}{x}}=x$, is not the domain $\mathbb{R}$?

Answer 1

I really like your argument. But the truth is, you could only make the simplification when $x$ is in the domain of $f(x)=\frac1x$ and of course $f(f(x))$. In this case, $f(0)$ doesn't exist, so the domain of $f(f(x))$ doesn't include $0$.

However, we do have that $$\lim_{x\rightarrow0}f(f(x))=0$$Because we could substitute $x=0$ after simplification.

Answer 2

As noticed it is absolutely true that

$$\frac{1}{x^{-1}}=\frac{1}{\frac{1}{x}}=x$$

and also that the domain for $x$ is $\mathbb{R}$ but your original expression is not defined at $x=0$ therefore its domain is exactly $x\in \mathbb{R}-\{0\}$.

In other words the two expressions are completely equivalent but only for $x\neq 0$.

This kind of discontinuity is defined as a removable one since we can set the value at $x=0$ for $\frac{1}{x^{-1}}$ equal to zero and in this way its domain becomes $\mathbb{R}$.