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The logarithm is not defined at $x=0$, because it tends to $-\infty$ as x tends to 0 from above. But is it nevetheless correct to say that $$ \log_2(0)=-\infty? $$

Or is it better to say/ write $$ \log_2(0)=\lim_{x\downarrow 0}\log_2(x)=-\infty $$

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    $\begingroup$ No.$\qquad\qquad$ $\endgroup$ Nov 30, 2014 at 1:30

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It is correct to say that $\lim_{x\to 0^+} \log(x) = -\infty$. That is, $\log x$ tends toward $-\infty$ as $x \to 0^+$.

However, $\log(0)$ is not equal to $-\infty$, because, as you say, $\log(x)$ is undefined at $x = 0$.

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  • $\begingroup$ What's with something like $\lim_{n\to\infty}\frac{1}{n}\log_2 x_n$ if $x_n=0\forall n\geq 1$? $\endgroup$
    – mathfemi
    Nov 27, 2014 at 13:45
  • $\begingroup$ That limit is indeterminate. $\endgroup$
    – amWhy
    Nov 27, 2014 at 13:48
  • $\begingroup$ Hm, but in a book is it said that this is $-\infty$. $\endgroup$
    – mathfemi
    Nov 27, 2014 at 13:57
  • $\begingroup$ See Wolfram here Note that indeterminate isn't equivalent to "undefined" nor to "limit does not exist." $\endgroup$
    – amWhy
    Nov 27, 2014 at 14:00
  • $\begingroup$ Ok. I am not familiar with this expression. How can the book then claim that it is $-\infty$? $\endgroup$
    – mathfemi
    Nov 27, 2014 at 14:03
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as you said, it is not defined at 0. But, (here I wave my hand), it is ok to think it is $-\infty$ when you work in limit sense.

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You could define an element called $-\infty$ and then define, for example, $\ln: \left [{0..+\infty}\right) \to \mathbb R\cup\{-\infty,+\infty\}$. There are maths that do this, and the codomain is called the "extended reals". But there's no point in doing so here because $-\infty$ wouldn't have the properties of regular numbers in terms of multiplication, subtraction etc. Also, it would be confusing, because you'd be using $\ln$ as a different function than other people. So don't do it.

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