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What is $\ln 0$ ?

Is it $-\infty$ or indeterminate?

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In a lot of ways, which one you use depends on which is more useful for the problem at hand. For limit operations, as long as you are taking the limit coming from above, $-\infty$ is useful. Otherwise, indeterminate is more useful. –  Ross Millikan Apr 29 '13 at 2:09
    
It's undefined. But $\lim_{x \to 0^+} \ln x = -\infty$. If you extend the domain of $\ln$ to $[0, \infty)$ and if you allow the extended function to take values in the "extended real numbers" $[-\infty, \infty]$ or just in $[-\infty, \infty)$, then you might be able to say that your modified $\ln$ satisfies $\ln 0 = -\infty$. Without any context indicating that this has been done, then $\ln 0$ is just undefined. –  Stefan Smith Apr 29 '13 at 23:23

2 Answers 2

up vote 12 down vote accepted

$\ln 0$ is undefined. Why is that? Remember that $y = \ln x$ is defined as the unique number staisfying $e^y = x$. But we know that the exponential function is always positive, so what happens if we take $x = 0$? Then there's no $y$ that will make the equation $e^y = 0$ true, so $\ln 0$ is undefined.

However, we can take limits, and in that context we can say $\lim\limits_{x \to 0^+} \ln x = -\infty$, because the closer $x$ gets to $0$ the lower the $y$ we need to take to make $e^y = x$ true. But this does not mean that $\ln 0 = -\infty$! Infinity is not a number and $\ln 0$ doesn't exist, so it doesn't make sense to use them in an equation like this.

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The expression $x = \ln 0$ is basically the equivalent to

$$e^x = 0$$

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