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Infinity question - limits

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  1. Jul 14, 2016 #1

    Rectifier

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    Is ## "\frac{0}{\infty}"=0 ## ?
     
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  3. Jul 14, 2016 #2

    phinds

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    No, it is undefined because you are treating infinity as though it was a number. It isn't. You can't, in most circumstances, use infinity in normal math and expect meaningful results.
     
  4. Jul 14, 2016 #3
    ## \infty ## isn't a number, so you shouldn't be dividing by it.
    That said, in most contexts, yes, ## \frac{0}{\infty}=0 ##
     
  5. Jul 14, 2016 #4

    phinds

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    No, it's still undefined in most contexts, I think. What IS true is that as n approaches infinity, 0/n approaches 0.

    EDIT: I hasten to add, I don't do any math where there IS any meaningful context for 0/infinity, so I could be wrong. What contexts did you have in mind?
     
  6. Jul 14, 2016 #5
    OK, in most meaningful contexts.
    As a matter of practicality, if someone ended up with infinity in the denominator, they were probably using shorthand in dealing with a limit. The type of nomenclature is commonly used to explain limits - with the caveat that it is not proper general-use nomenclature. Since the OP put the fraction in quotes, I believe the he already knows about this.

    The symbol ## \infty ## can be used in other more proper contexts, none of which would allow it to be in a denominator. One is to indicate the result of a limit. Another is to indicate the cardinality of a set, although Aleph is better at that. In cases where the order of Alephs is important, the omega symbol is used. Division can be defined for the omegas, but significant caveats are needed.
     
    Last edited: Jul 14, 2016
  7. Jul 14, 2016 #6

    Rectifier

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    I was trying to calculate following limit:
    ## \lim_{x\rightarrow \infty} \frac{x^4 + x \ln x}{x + \left( \frac{2}{3} \right)^x} = \lim_{x\rightarrow \infty} \frac{x^4 \left( 1 + \frac{\ln x}{x^3}\right) }{ x \left( 1 + \frac{ \left( \frac{2}{3} \right)^x }{x} \right) } = \lim_{x\rightarrow \infty} x^3\frac{1 + \frac{\ln x}{x^3} }{ 1 + \frac{ \left( \frac{2}{3} \right)^x }{x} } ##

    ## \lim_{x\rightarrow \infty} x^3 = \infty ##

    ## \lim_{x\rightarrow \infty}\frac{\ln x}{x^3}=0 ##

    ## \lim_{x\rightarrow \infty} \left( \frac{2}{3} \right)^x = 0 ##

    ## \lim_{x\rightarrow \infty} \frac{ \left( \frac{2}{3} \right)^x }{x}
    = "\frac{0}{\infty}"##

    HERE-----------------^
     
  8. Jul 14, 2016 #7

    Mark44

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    That last limit is 0. Don't write it as ##\frac 0 {\infty}##, either with or without quotes.

    BTW, I am moving this thread to the technical math section, as it is more of a conceptual question than an actual homework-type question.
     
  9. Jul 14, 2016 #8

    Rectifier

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    I don't understand why the last limit is 0. :,(
     
  10. Jul 14, 2016 #9

    Rectifier

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    I think I got it.


    ## \lim_{x\rightarrow \infty} \frac{ \left( \frac{2}{3} \right)^x }{x}
    = \lim_{x\rightarrow \infty} \frac{ 1 }{x} \left( \frac{2}{3} \right)^x ##

    ## \lim_{x\rightarrow \infty} \frac{ 1 }{x} = 0 ##

    ## \lim_{x\rightarrow \infty} \left( \frac{2}{3} \right)^x = 0 ##
     
  11. Jul 14, 2016 #10
    as for me, this notation is completely clear. If we have two functions ##f(x)\to 0## and ##g(x)\to \infty## as ##x\to a## then ##f/g\to 0##
     
  12. Jul 14, 2016 #11

    mfb

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    That approach is much better, because 0*0 is defined in the real numbers.
     
  13. Jul 14, 2016 #12

    micromass

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    You are wrong. In most contexts where the OP makes sense, it is indeed true that ##0/\infty = 0##.
     
  14. Jul 14, 2016 #13

    micromass

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    0*0 is undefined?
     
  15. Jul 14, 2016 #14

    mathman

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    Why do you think so? 0 is a number, and multiplying any number by 0 is 0.
     
  16. Jul 14, 2016 #15

    mfb

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    Huh? I said it is defined.
     
  17. Jul 14, 2016 #16

    phinds

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    micromass, than you for that correction, but I wonder if you could expand on it for me a bit? I'm confused as to how it is that "0/infinity = 0" is not treating infinity like a normal number
     
  18. Jul 15, 2016 #17

    mfb

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    If you treat infinity like a normal number, you should be allowed to multiply by it, but then you get ##0=0\cdot \infty## which is not well-defined any more.
     
  19. Jul 15, 2016 #18

    phinds

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    But that's my point. I've always been told that you CAN'T normally treat infinity like a normal number. Is this wrong?
     
  20. Jul 15, 2016 #19

    mfb

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    Correct. You can still perform some limited operations in most contexts, and 0/infinity=0 is one of them (e.g. for limits). That is not treating it as normal number, that is a specific rule for this operation.
     
  21. Jul 15, 2016 #20

    phinds

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    OK, thanks.
     
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