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Are $\log_1 1$ and $\log_0 0$ indeterminate forms?

Whenever I ask someone about these indeterminate forms, they deny by saying either $\log$ is neither defined at base $0$ nor at base $1$, or they say $\log$ is a function so these must not be included in fundamental indeterminates.

But, we know division by zero is not defined, yet $0/0$ is indeterminate; and many others. And, actually, $\log$ is more a binary-operator that is the inverse operation of power/exponent operator.

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    $\begingroup$ Implicitly, you are saying that you do not believe that $\log_0(x)$ and $\log_1(x)$ are undefined. If they are not undefined, what are their definitions? $\endgroup$
    – Xander Henderson
    Jan 19, 2019 at 19:40
  • $\begingroup$ @Xander Henderson i am saying division by zero is yet undefined, but it gets more complicated, when numerator is also zero $\endgroup$ Jan 19, 2019 at 19:43
  • $\begingroup$ No, it really doesn't get any more complicated. $\frac{0}{0}$ is an undefined mathematical expression. $\endgroup$
    – Xander Henderson
    Jan 19, 2019 at 19:44
  • $\begingroup$ that's definitely not true, 0/0 and other indeterminants are so important that whole branch of limits grew out of them, they can't be simply undefined $\endgroup$ Jan 19, 2019 at 19:50
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    $\begingroup$ thnx @Blue for help $\endgroup$ Jan 19, 2019 at 20:07

3 Answers 3

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I think that your difficulty comes from a confusion regarding what an "indeterminate form" is. Indeterminate forms show up in analysis via naive substitution when computing limits. For example, we might naively compute $$ \lim_{x\to 0} \frac{x^2}{x} = \frac{\lim_{x\to 0} x^2}{\lim_{x\to 0} x} = \frac{0}{0}. $$ Since this last expression is undefined, we might say that the limit is "indeterminate of the form $\frac{0}{0}$." When this kind of naive substitution leads to an undefined expression, it is necessary to be a bit more clever in the evaluation of the limit. In this case, $$ \lim_{x\to 0} \frac{x^2}{x} = \lim_{x\to 0} x = 0. $$ Techniques for working with indeterminate forms include results such as L'Hospital's rule, applying algebraic transformations, and so on.

In the case of "the logarithm base 0", $\log_0(x)$ is undefined. This expression doesn't make sense. If this expression were defined, then it must be equal to some number, say $y$. Then $$ \log_0(x) = y \implies x = 0^y = 0. $$ But $0^y = 0$ for any positive value of $y$. Hence the expression $\log_0(x)$ is not well defined, as there is a not a unique value of $y$ which gets the job done. On the other hand, we can consider limits of expressions of the form $\log_b(a)$ as $b$ tends to zero and $a$ either tends to zero or diverges to infinity. Such limits can be said to be "indeterminate of the form $\log_0(0)$" or "indeterminate of the form $\log_0(\infty)$, but this does not mean that they are equal to either of these expressions (anymore than $\lim_{x\to 0} x^2/x = 0/0$).

Such limits typically require more careful analysis, again using algebraic tools, L'Hospital's rule and other results from analysis, bounding with estimates, or direct $\varepsilon$-$\delta$ style computation. Limits involving logarithms are discussed in greater detail in J.G.'s answer.


In short, when we say that "the limit is indeterminate of the form $X$", we are saying that if we try to evaluate the limit by naive substitution, then we get the expression $X$, where $X$ is some undefined expression like $\frac{0}{0}$, $\log_0(0)$, or $1^\infty$. Such limits cannot be evaluated by naive substitution, and require other techniques.

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  • $\begingroup$ WOW! by the way i am struggling now i think $\endgroup$ Jan 19, 2019 at 20:15
  • $\begingroup$ so basically it needs more analysis $\endgroup$ Jan 19, 2019 at 20:16
  • $\begingroup$ all this means 0/0 is undefined $\endgroup$ Jan 19, 2019 at 20:18
  • $\begingroup$ thnx thnx thnx, really learnt something new $\endgroup$ Jan 19, 2019 at 20:19
  • $\begingroup$ :) wow really thank you, thnx for breaking the mirage $\endgroup$ Jan 19, 2019 at 20:19
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Suppose $f,\,g\to 1$ as $x\to 0$. How can we vary $\lim_{x\to 0}\log_f g$? We can take $f=\exp x,\,g=\exp cx$ to get any value $c\in\Bbb R$ we like, or to get $\pm\infty$ we can use $f=\exp x^2,\,g=\exp\pm 1$.

Suppose $f,\,g\to 0$ as $x\to 0$. How can we vary $\lim_{x\to 0}\log_f g$? We can take $f=x,\,g=x^c$ to get any value $c\in\Bbb R^+$ we like, or to get $\infty$ we can use $f=x,\,g=x^{1/x^2}$. We can't achieve negative limits because $\ln f,\,\ln g\to -\infty$.

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  • $\begingroup$ wow! i didnt even noticed that, that's amazing, i didnt thought about there ranges $\endgroup$ Jan 19, 2019 at 19:57
  • $\begingroup$ @PranshuKandal This limit-theoretic treatment explains all the indeterminate forms. $\endgroup$
    – J.G.
    Jan 19, 2019 at 19:58
  • $\begingroup$ thnx @J.G. for help $\endgroup$ Jan 19, 2019 at 20:05
  • $\begingroup$ got that as zero to the power of any non-positive number is not defined, log 0 base 0 can only be positive. $\endgroup$ Jan 19, 2019 at 20:11
  • $\begingroup$ @PranshuKhandal $\log_0(0)$ is undefined. If $f(x) \to 0$ and $g(x) \to 0$, then $\lim \log_{f(x)}(g(x))$ is (naively) "indeterminate of the form $\log_0(0)$", which may be nonnegative, infinite, or undefined. Because the limit is of an indeterminate form when evaluated naively, deeper analysis is required. $\endgroup$
    – Xander Henderson
    Jan 19, 2019 at 20:16
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Let $a=\log_1(1)$ and suppose $a$ exists, then $$a=\log_1(1)\iff 1^a=1$$ so $a$ can be any number. Similarly for $b=\log_0(0)\iff 0^b=0$

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  • $\begingroup$ does it not prove them indeterminant?? and yes, nice name you have got :) $\endgroup$ Jan 19, 2019 at 19:47
  • $\begingroup$ Indeterminant in the sense that any value works, both $a$ and $b\neq 0$ can take any value. $\endgroup$ Jan 19, 2019 at 19:51
  • $\begingroup$ can we add them into the list of our fundamental indeterminants $\endgroup$ Jan 19, 2019 at 19:53
  • $\begingroup$ if not? them why $\endgroup$ Jan 19, 2019 at 19:53
  • $\begingroup$ i think I've got the answer, but which do i accept, help me out $\endgroup$ Jan 19, 2019 at 19:55

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