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Homework Help: Derivable function

  1. May 2, 2006 #1
    Hey everyone
    I was going through some problems and I came across this one, it said to see if the fuction below is derivable in the point [tex]x_{0}=1^{-}[/tex]
    [tex]f(x)=\frac{x+1}{1-\sqrt{1-x}}[/tex]

    so I did this as usual,
    limit of [tex]\frac{f(x)-f(1)}{x-1}[/tex] when x -> 1, and x<1 is equal to the limit of:

    [tex]\frac{x-1+2\sqrt{1-x}}{(x-1)(1-\sqrt{1-x}}[/tex]


    but I really don't know what to do after!! I tried to multiply the top and bottom with "x+2sqrt(1-x)" but with no satisfying results.

    Any help would be appreciated!
     
    Last edited: May 2, 2006
  2. jcsd
  3. May 3, 2006 #2

    J77

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    What do you mean by "derivable"?

    Obviously, for [tex]x>1[/tex] you get complex solutions, and [tex]f(x)\rightarrow\pm\infty[/tex] as [tex]x\rightarrow0[/tex] (- from left, + from right)...

    I wrote it as:

    [tex]f(x)=\frac{(x+1)(1+\sqrt{1-x})}{x}[/tex]
     
  4. May 3, 2006 #3
    Maybe I meant differentiable... sorry!
    what I need to find the limit of is this function [tex]f(x)=\frac{x-1+2\sqrt{1-x}}{(x-1)(1-\sqrt{1-x})}[/tex] as [tex]x\rightarrow1^{-}[/tex]
    Thank you
     
  5. May 3, 2006 #4

    VietDao29

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    Wait, can you copy exactly what the problem says please. I dunno, but f(1) is undefined, so the function is not differentiable at x = 1.
     
  6. May 3, 2006 #5

    J77

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    Why undefined?

    Because [tex]\sqrt{0}[/tex]?

    As you approach from the left, it tends to 2 tho'...

    (and the derivative goes to [tex]\infty[/tex]?)
     
    Last edited: May 3, 2006
  7. May 3, 2006 #6

    Curious3141

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    The limit of that expression is easy enough to find. Just substitute [tex]u = \sqrt{1-x}[/tex]

    But I don't know what your question is asking, or, frankly, what you're trying to accomplish on the whole. Post the question verbatim, please.
     
  8. May 3, 2006 #7
    [tex]f(x)=\frac{x-1}{(x-1)(1-\sqrt{1-x})}+\frac{2}{1-x-\sqrt{1-x}}[/tex]
    the first one's lim=1 and second one's lim=0
     
  9. May 3, 2006 #8

    VietDao29

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    Ack, ack, doing maths late is never good... :cry: :cry: :cry:
    Okay, for mohlam12's question, have you considered factoring the [tex]\sqrt{1 - x}[/tex] out in the numerator?
    [tex]\lim_{x \rightarrow 1 ^ -} \frac{x - 1 + 2 \sqrt{1 - x}}{(x - 1) (1 - \sqrt{1 - x})} = \lim_{x \rightarrow 1 ^ -} \frac{\sqrt{1 - x} (\sqrt{1 - x} - 2)}{(1 - x) (1 - \sqrt{1 - x})} = ...[/tex]
    You can go from here, right? :)
    Sorry for such confusion... My bad :frown:
     
  10. May 3, 2006 #9
    Yup, I can go from here, thanks. I just didn't have that idea to factor with [tex]\sqrt{1-x}[/tex]
     
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