Is this function defined at zero?

Question

In the sense of real analysis, is this function defined at zero?

$$f(x) = \frac{1}{\frac{1}{x}} = (x^{-1})^{-1}$$

Answer 1

Note that the multiplicative inverse of a real number $x$, denoted $x^{-1}$, is defined such that $x\cdot x^{-1} = 1$. There is the requirement that $0$ has no multiplicative inverse (otherwise we will have sacrificed some arithmetic nice properties). So we write $x^{-1} (= 1/x)$ if and only if $x \neq 0$. So the answer to your question is no.

It is very tempting to write $(x^{-1})^{-1} = x$ and to think that this applies to whatever real $x$ without recalling first that that we can write $x^{-1}$ for and only for $x \neq 0$.

Answer 2

No it isnt, as without simplifying, you can't "straightfoward" plug in $0$. Simplifying assumes the denominator isn't zero and you get $f(x)=x$ for all $x\neq0$. It's worth noting the function is tending to zero as $x\rightarrow0$ though.