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Is it possible to let limit multiplication rule not be exist?

  1. Oct 8, 2012 #1
    We have known that:
    lim f(x)g(x) = lim f(x) * lim g(x)

    Could any conditions or limitations make: lim f(x)g(x) ≠ lim f(x) * lim g(x)?
  2. jcsd
  3. Oct 8, 2012 #2


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    Writing out the proof really quickly, I don't see what kind of limitation or conditions can make this not true. Perhaps if you proved it, you would see why this is always true?

    Edit: Based on micro post, I did the terrible mistake of assuming you meant that the x approached the same a and both limits exist.
    Last edited: Oct 8, 2012
  4. Oct 8, 2012 #3


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    No, this is not known. And this is not even true. You need to be very careful with statements like this. The correct statement is that

    [tex]\lim_{x\rightarrow a} f(x)g(x)=\lim_{x\rightarrow a} f(x) \lim_{x\rightarrow a} g(x)[/tex]

    if both limits [itex]\lim_{x\rightarrow a} f(x)[/itex]and [itex]\lim_{x\rightarrow a} g(x)[/itex] exist and equal a real number.

    If we allow the limits to become infinite, then the rule does not hold anymore. For example, it is not true that

    [tex]\lim_{x\rightarrow 0} \frac{x^2}{x^2}=\lim_{x\rightarrow 0} x^2 \lim_{x\rightarrow 0}\frac{1}{x^2}[/tex]

    since the limit [itex]\lim_{x\rightarrow 0}\frac{1}{x^2}[/itex] is not a real number (but rather [itex]+\infty[/itex]).

    You need to be very careful in mathematics to always give the exact statements and conditions in which something holds.
  5. Oct 8, 2012 #4
    Big Thanks to micromass and MarneMath, both of you bestead me a lot.
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