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Defining f(0) to be continuous

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    f(0,1) ---> R by f(x) =1/x^(1/2) -((x+1)/x)^(1/2). Can one define f(0) to make f continuous at 0?


    2. Relevant equations
    lx-x0l<delta
    lf(x)-f(x0l<epsilon


    3. The attempt at a solution
    My thought is that the limit must equal f(0), but I'm unsure of how to get f(0) because of division by zero.
     
  2. jcsd
  3. Nov 15, 2008 #2

    arildno

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    Write this as a single fraction:
    [tex]f(x)=\frac{1-\sqrt{x+1}}{\sqrt{x}}[/tex]
    Now, utilize the following identity in a constructive fashion:
    [tex]1=\frac{1+\sqrt{x+1}}{1+\sqrt{x+1}}[/tex]
     
  4. Nov 15, 2008 #3
    ok, so [tex]\frac{x-x^{2}}{\sqrt{x+1}}[/tex]
     
  5. Nov 15, 2008 #4

    HallsofIvy

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    And? Since f(x) and that are exactly the same for all x except x= 0, they have the same limit at x=0. What is the limit as x goes to 0? Define f(0) to be that limit.
     
  6. Nov 15, 2008 #5
    Ok, so the limit = 0. Let f(0)=0?
     
  7. Nov 15, 2008 #6

    HallsofIvy

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    Yes, of course!
     
  8. Nov 15, 2008 #7
    Ok thanks!
     
  9. Nov 16, 2008 #8
    Ok, I think I did this wrong as I look at it today. Can anybody help me simplify a bit?
     
  10. Nov 16, 2008 #9

    HallsofIvy

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    You are multiplying numerator and denominator of
    [tex]f(x)=\frac{1-\sqrt{x+1}}{\sqrt{x}}[/tex]
    by [tex]1+ \sqrt{x+1}[/tex] so then numerator becomes [tex]1- (\sqrt{x+1})^2= x[/itex] and the denominator is [tex]\sqrt{x}(1+ \sqrt{x+1})[/tex]. Now, [tex]x/\sqrt{x}[/tex] is [tex]\sqrt{x}[/tex] so the fraction becomes
    [tex]\frac{\sqrt{x}}{1+ \sqrt{x+1}}[/tex]
    What is the limit of that as x goes to 0?
     
  11. Nov 16, 2008 #10

    arildno

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    The numerator is -x, Halls, not x!
     
  12. Nov 16, 2008 #11
    [tex]1- (\sqrt{x+1})^2= x[/itex]
    Ok, I don't see this. isn't it just -x because 1-(x+1)=-x?
     
  13. Nov 16, 2008 #12
    Limit = 0
     
  14. Nov 16, 2008 #13

    arildno

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    HallsofIvy just forgot the minus sign.
     
  15. Nov 16, 2008 #14
    Ok, i thought that was just it.
     
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