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Need help with limits

  1. Apr 4, 2014 #1
    1. The problem statement, all variables and given/known data


    Suppose the lim x-> a f(x) = + infinity and lim x->a g(x) = 0

    then why would lim x→a (f(x)×g(x)) be insufficient to tell us anything about the product of 2 limit?
     
  2. jcsd
  3. Apr 4, 2014 #2
    Because the product of ##\infty \times 0## is not defined.

    Write ##g(x)f(x)=\frac{g(x)}{\frac{1}{f(x)}}## and now calculate the limit.
     
  4. Apr 4, 2014 #3

    Ray Vickson

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    Show your work. Where are you stuck?
     
  5. Apr 4, 2014 #4

    LCKurtz

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    Look at these two examples taking ##a=0## so ##x\rightarrow 0^+##:

    ##f(x) = \frac 1 x \rightarrow \infty,~g(x) = x^2\rightarrow 0##. Here ##f(x)g(x) = x\rightarrow 0##

    Now take ##f(x) = \frac 1 {x^2}\rightarrow \infty,~g(x) = x\rightarrow 0##. Here ##f(x)g(x)= \frac 1 x\rightarrow \infty##.

    So you having ##f\to\infty,~g\to 0## isn't sufficient to tell us anything about ##fg##.
     
    Last edited: Apr 4, 2014
  6. Apr 4, 2014 #5

    HallsofIvy

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    And to complete what LCKurtz said, for a any non-zero number,
    If [itex]f(x)= \frac{a}{x}[/itex] and [itex]g(x)= x[/itex], then [itex]\lim_{x\to 0} f(x)g(x)= a[/itex]

    So that, in fact, there are examples giving every possible result!
     
  7. Apr 5, 2014 #6
    This is tough to grasp or maybe I'm missing some intermediate steps.

    By the basic limit law of multiplication:

    lim x->a f(x) and lim x->a g(x)
    then lim x->a f(x) . lim x->a g(x) = lim x->a f(x) .g(x)

    so f(x) -> infinity and g(x) -> f(x).g(x) = 0
     
  8. Apr 5, 2014 #7

    Mark44

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    Equal what?

    From post #1, ##lim_{x \to a} f(x) = \infty## and ##lim_{x \to a} g(x) = 0##.
    The property you are citing, about the multiplication of limits, requires that both limits exist. That means that each limit has to be a finite number. So your first limit does not exist.
    No.
    There are already several examples in this thread that show that this is not a valid conclusion.
     
  9. Apr 5, 2014 #8
    Why complicate things by mixing up a and zero?
     
  10. Apr 5, 2014 #9

    HallsofIvy

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    I don't know what you mean by this. Mixing up "a" and what "zero"?
     
  11. Apr 5, 2014 #10

    Mark44

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    HallsOfIvy is making the point that an [∞ * 0] indeterminate form can turn out to be any number.
     
  12. Apr 5, 2014 #11
    I believe there was only one zero in your post - that zero was the one you apparently don't see as the problem.

    The OP's question was to do with the limit as [itex]x[/itex] tended to [itex]a[/itex]. Your comment was to do with [itex]x[/itex] tending to [itex]0[/itex] with an [itex]a[/itex] in the expression - a very different [itex]a[/itex]!

    It may be considered to be a small pedantic point, but I think it's very important and to my mind you added unnecessary confusion. Perhaps your mind was swayed by LCKurtz's example of [itex]a[/itex] being [itex]0[/itex].

    Thanks, I was aware of that. :smile:
     
    Last edited: Apr 5, 2014
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