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There maybe a mistake in the question but, let that someone asks you to calculate something like this: $$0\cdot \lim_{x\to0}(log(x))$$ with no further information. The assumptions that one makes is just that log is the natural logarithm, $x\epsilon\mathcal{R}$ and generally maybe some assumptions that a first year calculus course would assume. Nothing too complicated.

The question is the following: How do you somewhat rigorously attack this thing?

My thoughts:

If you see this as a whole is an undefined quantity of the type: $0 \cdot \infty$.

If you see it as parts you have a number $0$ and a limit that diverges. Since the limit does not exist (of course we implicitly assume that $\lim_{x\to0^{+}}$) we could not use the multiplication rule.

Hypothetically if we could use the multiplication rule we run into problems of what function's limit should we represent $0$ with. $x$? $x^2$? $x^{1/10}$?

What do I say then about this object? Does it even make sense to ask something like that?

jmstf94
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    Usually the question is about $ \lim_{x\to0}f(x)(\log(x))$ where $ \lim_{x\to0}f(x)=0$. – Dietrich Burde Jun 17 '20 at 08:39
  • Well the expression as you write is meaningless. Technically $\lim_{x\to 0^{+}}\log x$ is not something which can be manipulated like real numbers. Compare this with $\lim_{x\to 1}\log x$ which can be manipulated like numbers because this is just $0$. – Paramanand Singh Jun 17 '20 at 10:09

2 Answers2

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An algebraic expression needs to have all its terms defined to have a meaning.

As

$$\lim_{x\to0}\log x$$ is undefined, the whole expression $0\cdot\lim_{x\to0}\log x$ is undefined.

And we also have $$(\lim_{x\to0} x)(\lim_{x\to0}\log x)$$

undefined, while

$$\lim_{x\to0}(x\log x)=0.$$

Also note that $0\cdot\infty$ is not an expression but an expression pattern which describes a limit of the form

$$\lim_{x\to a}(f(x)g(x))$$ where

$$\lim_{x\to a}f(x)=0\text{ and }\lim_{x\to a}g(x)=\infty,$$ as in my third example only.

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Generally $\lim_{x\to x_0}f(x)$ is not to be taken as a number. This thing has no meaning by itself. The only meaningful expression is $\lim_{x\to x_0}f(x)=y$, as a whole. Here $y\in\mathbb R \cup\{\pm\infty, \text{DNE}\}$. And $a \lim f(x)$ actually means "We already know that $\lim f(x)=c$ is a number, and we want to look at $ac$," plus some convenient edge cases involving $\pm \infty$. So your expression really has no meaning at all.

Trebor
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    I don't quite agree. You might very well write $f(\lim_{x\to x_0}x)$. What is a special convention is to write $\cdots=\infty$ (as well as $\cdots\to\infty$). –  Jun 17 '20 at 09:18
  • @YvesDaoust Well then I guess it depends on the convention. My textbook unambiguously stated this at the introduction of limits. Anyway these definitions get you the same result ;) – Trebor Jun 17 '20 at 09:21
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    Do you mean that $f(\lim_{x\to x_0}x)$ is not allowed ? –  Jun 17 '20 at 09:22
  • @YvesDaoust As I said, it is a convenient shorthand for $c = \lim blah$ and then $f(c)$. – Trebor Jun 17 '20 at 09:23
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    @Trebor, you misinterpreted something, because, otherwise, we couldn't talk about continuous functions the way we do, at all. If $\lim\limits_{x\to x_0} f(x)$ exists, then it is a number... Event if it doesn't exist, $f\left(\lim\limits_{x\to x_0} x\right)$ is perfectly fine. – PinkyWay Jun 17 '20 at 10:17
  • @Croissant Well, I can just pass off every argument you have as being a "convenient shorthand", because convenient shorthands are exactly how we come to talk about anything the way we do, at all. For instance, the notation $c \in {a,b}$ is a (very) convenient shorthand for a tremendous amout of reasoning that certain sets exists, and $c$ belongs to them... Am I clear enough? – Trebor Jun 17 '20 at 10:21
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    No, you aren't. What if $f$ is a multivariable function? Its limit, if exists, definitely isn't a number, but an ordered $n$ tuple... – PinkyWay Jun 17 '20 at 10:22
  • @Croissant Since my interpretation allows almost everything yours does, I think it is pointless for extended arguments here. – Trebor Jun 17 '20 at 10:27
  • @Croissant Well, you should not edit your comments except for correcting typos, because I will not receive notifs that way. In fact, many books on mathematical logic ban the use of functions altogether. The only legitimate thing is to say $f(x)=y$, as a whole. $f(g(x))=y$ is not legal, and should be translated to $g(x)=z, f(z)=y$. You can see an instance of this in Cohen's Set Theory and the Continuum Hypothesis. It is worth emphasizing that my approach is as strong as yours. It doesn't "exclude" interesting mathematics. So arguing over this is really worthless. – Trebor Jun 17 '20 at 12:14