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As per the title; what is $ \lim_{x \to 0} \log_0(x) $ ?

According to WolframAlpha: $$ \lim_{x \to 0} \log_0(x) = 0 $$ but how is this possible?

Surely the limit should be indeterminate since $\log_0(x) = \frac{\log(x)}{\log(0)} $ and $ \log(0) = $ indeterminate?

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As $X\to 0$ , $ln(X)\to-\infty$. So now we use the change of base formula, $log_a(X)=\frac{ln(X)}{ln(a)}$, thus $lim_{a\to 0}log_a(X)=lim_{a\to 0}\frac{ln(X)}{ln(a)}=0$ –  Baby Dragon Mar 20 '13 at 2:37
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What does $\log_0(x)$ mean? You have to have this make sense before you can work on this limit. –  alex.jordan Mar 20 '13 at 2:39
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If I understand you, then the limit would be $\lim\limits_{x\to0}\log_x(x)$, not $\lim\limits_{x\to0}\log_0(x)$. –  alex.jordan Mar 20 '13 at 2:42
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$\log_0(x)$ asks the question "what must we raise $0$ to, to get $x$?" So $\log_0(x)$ is undefined everywhere, except by many people's convention, at $x=1$, where $\log_0(1)=0$. –  alex.jordan Mar 20 '13 at 2:43
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Log to the base $0$? Got to be kidding. It may be interesting (though, it seems, not particularly challenging) to get Alpha to do something stupid. –  André Nicolas Mar 20 '13 at 2:46

1 Answer 1

up vote 3 down vote accepted

$\lim\limits_{x\to0}\log_0 x$ cannot exist unless $\log_0 x$ exists for $x$ in some open neighborhood of $0$, with the possible exception of $x=0$. (Since the base must be positive, we can take "open neighborhood of $0$" to mean a set of the form $[0,\varepsilon)$ where $\varepsilon>0$.)

Later edit: I suspect what's going on is something like this: $$ \log_\varepsilon x = \frac{\log x}{\log\varepsilon}, $$ then letting $\varepsilon\downarrow0$, we have the denominator going to $-\infty$, so that the fraction approaches $0$. Hence $\log_0 x$ get construed to be $0$, and then one lets $x$ go to $0$, and the limit is $0$.

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The function $\log_0(X)$ is continuous on it's very limited domain. –  Baby Dragon Mar 20 '13 at 2:46
    
But it's very limited domain is far from $0$. –  alex.jordan Mar 20 '13 at 2:49
    
@BabyDragon : Its very limited domain is empty, and it is indeed continuous at every point in its domain. But could you write $\log_0$ instead of $log_0$? Just use a backslash, thus: \log_0. This not only prevents italicization, but also provides proper spacing before and after "$\log$" in expressions like $a\log b$. Similarly \sin, \cos, \min, \max, \det, \inf, \sup, \lim, etc. –  Michael Hardy Mar 20 '13 at 2:49
    
I was referring to the part of the statement where you say "with the possible exception of $x=0$", if we choose to take this convention. –  Baby Dragon Mar 20 '13 at 2:52
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@MichaelHardy: Regarding $0^0=1$, see for example Knuth's "Two Notes on Notation", Amer. Math. Monthly, May 1992 (in particular the top of p. 408). –  Hans Lundmark Mar 20 '13 at 8:03

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