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Limit of a function

  1. Dec 11, 2008 #1
    For -1 < x < 1, x[tex]\neq[/tex]0 let
    f(x)= [tex]\frac{\sqrt{1+x}-1}{x}[/tex]

    i. Prove that
    lim f(x)
    x[tex]\rightarrow[/tex]0
    exists and find it
    (There is an easy proof and you get no credit for applying "hospital's rule")
    Part (i) shows that f can be continued to a continuous function on (-1,1) if we assign this limit to be f(0) (this is assumed in subseguent parts)

    ii. Show that the limit
    lim 1/x [tex]\int[/tex] f(t) dt
    x[tex]\rightarrow[/tex]0
    exists

    iii. Determine constants a0, a1, a2, a3 so that
    [tex]\int[/tex] f(t) dt = a0 + a1x + a2x2 + a3x3 + x3 [tex]\rho[/tex](x)
    (this integral goes from... 0 to x i couldn't figure out a way to put it on the integral.)
    where
    lim [tex]\rho[/tex](x) =0
    x[tex]\rightarrow[/tex]0


    I don't know how to go about this at all.

    Please help me out
    Thanks
     
  2. jcsd
  3. Dec 11, 2008 #2

    gabbagabbahey

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    Start with part (i). What must be true of [tex]\lim_{x\to0^{+}} f(x)[/tex] and [tex]\lim_{x\to0^{-}} f(x)[/tex] for [tex]\lim_{x\to0} f(x)[/tex] to exist?
     
  4. Dec 11, 2008 #3
    if both the integral approaches the same constant from the negative and positive side, then the function will approach that constant.

    but, where do i go from there?
     
  5. Dec 11, 2008 #4

    gabbagabbahey

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    Integral; what integral? :confused: Don't you mean 'limit'?

    For the limit to exist, both one-sided limits must approach the same value.

    That means that if both one-sided limits approach the same value, then the limit exists.

    So.....do both one-sided limits approach the same value?
     
    Last edited: Dec 11, 2008
  6. Dec 11, 2008 #5

    HallsofIvy

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    For (i) you might want to rationalize the numerator.

    For (ii) remember that an integral is always differentiable. What is the "difference quotient" for this?

    For (iii) I don't understand what is wanted if f itself is not given explicitely.
     
  7. Dec 11, 2008 #6
    I assume f means the function of (i) (continued continuously at zero)
     
  8. Dec 11, 2008 #7

    HallsofIvy

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    Oh, yes, I see now. i, ii, and iii all come after the definition of f. But no integration is necessary for ii.
     
  9. Dec 11, 2008 #8

    HallsofIvy

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    tomboi03, think about a Taylor's series expansion.
     
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