Is $\frac1{\log x}$ defined at $x = 0$?

Question

Since $\log (x)$ is only defined for $x \gt 0$, then $1/\log (x)$ should only be defined for $x \gt 0$.

However, my graphing calculator says otherwise.

Similarly, would $x\log(x)$, $x/\log (x)$ and $x + \log (x)$ be defined at $x = 0$?

Answer 1

$\log (0)$ doesn't exists in any case. And, yes, $\log x$ is well-defined for $x>0$. However, the function $$f(x) = \frac{1}{\log x}$$ is defined for $x>0$ and for the $x$'s such that $\log x \neq 0$, that is, for $x\neq 1$.

Answer 2

In fact you calculator use the analytic continuation for $f(x)=\dfrac1{\log x}$ as $$f(x)= \begin{cases} \dfrac{1}{\log x}\text{ if } x>0\\ 0\text{ if } x=0 \end{cases}$$

Answer 3

• $\lim_{x\to0^+} \frac{1}{\log x} = 0^-$
• $\lim_{x\to0^+} x{\log x} = 0^-$
• $\lim_{x\to0^+} \frac{x}{\log x} = 0^-$
• $\lim_{x\to0^+} x+\frac{1}{\log x} = 0^+$

so you can extend the function even if not defined originaly.