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A constant can be moved through limit sign-any intuitive way to understand this?

  1. Jan 21, 2010 #1
    A constant can be moved through limit sign--any intuitive way to understand this?

    We know that a constant can be moved through limit sign. However, according to my understanding, this result follows from the theorem that the limit of a product is equal to the product of the limits, and when one of the multiplicand of the product is a constant, then if we take the limit of the constant, it will equal to the constant itself.

    But is there an intuitive or graphical way of showing that a constant can be moved through a limit sign?
     
  2. jcsd
  3. Jan 21, 2010 #2
    Re: A constant can be moved through limit sign--any intuitive way to understand this?

    Suppose we were considering the following:
    [tex]
    lim_{x \rightarrow a} \ c f(x)
    [/tex]
    Where c is a constant.

    Then, we can agree that
    [tex]
    c f(x) = \underbrace{f(x) + f(x) + ... + f(x)}_{c}
    [/tex]

    Thus,
    [tex]
    lim_{x \rightarrow a} \ c f(x) = lim_{x \rightarrow a} \underbrace{f(x) + f(x) + ... + f(x)}_{c}
    [/tex]

    Would you be willing to believe the sum of the limits is the limit of the sum? If so, we're done.
     
  4. Jan 22, 2010 #3

    HallsofIvy

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    Science Advisor

    Re: A constant can be moved through limit sign--any intuitive way to understand this?

    l'Hôpital, are you asserting that this is only true if c is a positive integer? That's the only case in which
    [tex]c f(x) = \underbrace{f(x) + f(x) + ... + f(x)}_{c}[/tex]

    (Added- Ah, I see, Juwant asked for an "intuitive" way of seeing it.)

    From the definition of "limit":
    If [itex]\lim_{x\to a} cf(x)= L[/itex], c a (non zero) constant, then given any [itex]\epsilon> 0[/itex], there exist [itex]\delta> 0[/itex] such that if [itex]|x-a|<\delta[/itex], then [itex]|cf(x)- L|< \epsilon[/itex]. Therefore, given any [itex]\epsilon> 0[/itex], [itex]|c|\epsilon[/itex] is also greater than 0 and there exist [itex]\delta> 0[/itex] such that if [itex]|x- a|< \delta[/itex], [itex]|cf(x)- L|< |c|\epsilon[/itex]. Now, [itex]|c||f(x)- L/c|< |c|\epsilon[/itex] so [itex]|f(x)- L/c|< \epsilon[/itex] so [itex]\lim_{x\to a} f(x)= L/c[/itex].

    That is, if [itex]\lim_{x\to a} cf(x)= L[/itex], then [itex]\lim_{x\to a}f(x)= L/c[/itex] which is the same as [itex]c \lim_{x\to a}f(x)= L[/itex].

    If c= 0, then cf(x)= 0 for all x so [itex]\lim_{x\to a} cf(x)= 0= 0(\lim_{x\to a} f(x))[/itex] as well.
     
    Last edited by a moderator: Jan 22, 2010
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