Why is $\frac{1}{\ln(0)} = 0$?

I thought $ln(0)$ is undefined.

The context is, I am looking for discontinuities in a function, and I expected $x=0$ to be a discontinuity since $ln(0)$ is undefined. Here's the relevant portion of the function.


I'm thinking that since $ln(0)$ is undefined that $x=0$. Therefore there is a discontinuity. Desmos says $f(0)=0$.

  • 4
    $\begingroup$ The function $f(x)=\frac{1}{\ln(x)}$ is indeed discontinuous at $x=0$. Showing the original function and context of your question may help us to answer your actual question though. $\endgroup$
    – ndhanson3
    Sep 25, 2021 at 20:53
  • 6
    $\begingroup$ Maybe it's talking about the limit of it when $x$ goes to $0$ from right. $\endgroup$
    – Emad
    Sep 25, 2021 at 20:54
  • 6
    $\begingroup$ You can define $f(0)=0$ and it will be continuous, with that said $f$ isn't defined on $x=0$ and $\ln(0)$ is not a thing. $\endgroup$
    – kingW3
    Sep 25, 2021 at 21:05
  • 2
    $\begingroup$ Desmos also thinks that $10^{100}+2-10^{100} = 0$ desmos.com/calculator/lgzmcvfqzp , so you have to be careful with any calculator that you use. $\endgroup$
    – TomKern
    Sep 26, 2021 at 13:40

1 Answer 1


$\ln0$ is indeed undefined . As for the function, $\frac{1}{\ln|x|+4}$ would also consequently be undefined at $x=0$, unless, as kingW3 says in the comments, you separately define it to be some value at 0.

I'm not clear on why you asked about $\frac{1}{\ln0}$, though, since that isn't what the original function is, and thus may not share discontinuities with it.

About the Desmos result, calculators tend to approximate values sometimes, so you might not want to trust them too much.


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