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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 $x = 0$?

thomasb
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Guy Hoka
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  • $1/\log(x)$ is not defined for $x=0$ but $\lim\limits_{x\to0^+}1/\log(x)=0$ – J. W. Tanner Sep 18 '19 at 21:18
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    Are you familiar with limits? If a function has a limit (even when it's undefined), it will appear to have a value when graphed. In this case, $\lim_{x \to 0^+} 1 / \log(x) = 0$, so the function will appear to have the value $0$ at $x = 0$ when graphed. – Theo Bendit Sep 18 '19 at 21:23

3 Answers3

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$\log (0)$ 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 $x\neq 1$.

azif00
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In fact you calculator use the analytic continuation for $f(x)=\dfrac1{\log x}$ as $$f(x)= \begin{cases} \dfrac{1}{\log x}\text{ if } x>0\\ 0\text{ if } x=0 \end{cases}$$

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

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

thomasb
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