1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    Staff Emeritus
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Sqrt limit question
Loading...