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$-\infty$ and $+\infty$ are two mathematical objects that we attach to the real number system to extend it. These objects are governed by a set of properties. However, I'm confused on a few points. Any clarification on these points would be greatly appreciated.

$(i)$ Are $\frac{1}{\infty}$ and $\frac{1}{-\infty}$ indeterminate$?$ Or they are equal to $0?$

$(ii)$ Is $\left(+\infty\right)^{\left(+\infty\right)}$ indeterminate or $+\infty?$

$(iii)$ For $-\infty<x<0,$ $\left(+\infty\right)^x=?$

$(iv)$ $\left(+\infty\right)^{\left(-\infty\right)}=?$

$(v)$ $0^{\left(+\infty\right)}=?$

$(vi)$ For $-\infty<x<0,$ $x^{\left(+\infty\right)}=?$

$(vii)$ $\left(-\infty\right)^{\left(+\infty\right)}=?$

Thanks in advance!

José Carlos Santos
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    I would say that all such expressions are best avoided. – Angina Seng Jul 22 '17 at 11:42
  • @LordSharktheUnknown I'd still like to know what they are? Indeterminate? Or we assign some value to it? –  Jul 22 '17 at 11:43
  • (i)=0, (ii)=$+\infty$, (iii) = $+\infty$ for $x>0$, $=0$ for $x<0$ and undefined for $x=0$. (iv) $=0$. (v) $=0$. (vi) $=+\infty$ for $x>1$, $=0$ for $0<x<1$, $=1$ for $x=1$, undefined for $x<0$. (vii) undefined. – Hellen Jul 22 '17 at 11:46
  • "Indeterminate" only has meaning in the context of limits. – hmakholm left over Monica Jul 22 '17 at 12:07
  • @HenningMakholm I understand. But how do I understand that an expression is indeterminate? For example in limiting argument (as in comment below Jose's answer, we cannot decide whether $0 \times \infty = 0$ or $\infty$.. –  Jul 22 '17 at 12:18
  • @Dragon: That is easy: An expression is never "indeterminate". The only thing the word "indeterminate" applies to is forms of limits, which is a different thing from an expression. A limit of the form "$0\times \infty$" (which really ought to be written as something like $({\to}0)\times({\to}\infty)$ such that this confusion is avoided) has an indeterminate form. But $0\times\infty$ is not an expression because $\infty$ is not a thing that you can do arithmetic on. – hmakholm left over Monica Jul 22 '17 at 12:23
  • Ah I see! Thanks. –  Jul 22 '17 at 12:34

1 Answers1

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  1. They are equal to $0$.
  2. It is $+\infty$.
  3. $0$.
  4. $0$.
  5. $0$.
  6. It's indeterminate.
  7. It has no meaning.
José Carlos Santos
  • 427,504
  • Thank you for the answer! Can you give me some source where I can read about these term and why they are assigned the corresponding values? –  Jul 22 '17 at 11:46
  • @Dragon He cannot even compute them properly. – Hellen Jul 22 '17 at 11:47
  • @Hellen So these are sort of axioms that come with the objects $-\infty$ and $+\infty ?$ $($i.e. one cannot derive them from a smaller set of axioms$)$ –  Jul 22 '17 at 11:50
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    @Dragon No. I was thinking in terms of limits. Consider the first point, for instance. If a sequence $(x_n){n\in\mathbb N}$ converges to $\pm\infty$, then $(1/x_n){n\in\mathbb N}$ converges to $0$. I can justify my other answers with the same type of argument. – José Carlos Santos Jul 22 '17 at 11:51
  • Okay. That's a nice argument. So in this line, $0 \times \left(+\infty\right) = 0$. $($Take $x_n=0$ and $y_n=n)$ for instance. On the other hand, taking $x_n=\frac{1}{n}$ and $y_n=n^2$ gives $0 \times \left(+\infty\right)=+\infty$. That seems odd. So what really is the value of $0 \times \left(+\infty\right),$ if it is not indeterminate? –  Jul 22 '17 at 11:56
  • @Dragon Usually $0$ is understood as the 'normal number' zero, whereas only $\infty$ and $-\infty$ are understood as limits. – SvanN Jul 22 '17 at 12:07
  • @JoséCarlosSantos how do you say $0^{+\infty}$ is $0$? Take $a_n=((-1)^n\frac{1}{n})^{n/2}$, $n=1,2,\dots$. Is this sequence goes to $0$? – MAN-MADE Jul 22 '17 at 12:31
  • @Dragon No. $0\times(+\infty)$ is indeterminate. Note that I used arbitrary sequences, whereas you used specifc ones. From your approach, all that can be deduced is that $0\times(+\infty)$ either is indeterminate or it is equal to $0$. – José Carlos Santos Jul 22 '17 at 12:32
  • I understand. Thanks! –  Jul 22 '17 at 12:34
  • @MANMAID If $\lim_{n\in\mathbb N}x_n=0$ and $\lim_{n\in\mathbb N}y_n=+\infty$, then $\lim_{n\in\mathbb N}{x_n}^{y_n}=0$. Of course, in order for this to make sense, one must assume that $(\forall n\in{\mathbb N}):x_N\geqslant0$. – José Carlos Santos Jul 22 '17 at 12:35