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Homework Help: Sqrt limit question

  1. Feb 12, 2009 #1
    i need to solve this limit
    [tex]
    \lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)
    [/tex]
    i tried
    [tex]
    \lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)=\\
    \lim_{x->\infty}\left (\frac{\frac{1}{\sqrt{x}}\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}}{\frac{1}{\sqrt{x}}} \right)
    [/tex]
    but i get 0/0

    ??
     
    Last edited: Feb 13, 2009
  2. jcsd
  3. Feb 12, 2009 #2

    Mark44

    Staff: Mentor

    Well, 0/0 is not an answer. Put some numbers in and see what you get. That will at least give you an idea of what the limit might be.
     
  4. Feb 13, 2009 #3
    i agree that 0/0 is not an answer
    how to solve it?
     
  5. Feb 13, 2009 #4

    HallsofIvy

    User Avatar
    Science Advisor

    That looks to me like a candidate for "rationalizing" Write it as
    [tex]\frac{\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)}{1}[/tex]
    and multiply both numerator and denominator by
    [tex]\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}\right)[/tex]

    You will get
    [tex]\frac{x+ \sqrt{x+\sqrt{x}}- x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}[/itex]

    Now use the standard "trick" when x is going to infinity: divide both numerator and denominator by the highest power of x, here [itex]\sqrt{x}[/itex], so every x is moved to the denominator:
    [tex]\frac{\sqrt{1+ \sqrt{1/x}}}{\sqrt{1+ \sqrt{(1/x)+ \sqrt{1/x^2}}}+ 1}[/tex]

    As x goes to infinity, each of those fractions goes to 0.
     
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