Results 1 to 3 of 3

Math Help - sqrt limit question..

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    sqrt limit question..

    i need to solve this limit
    <br />
\lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)<br />
    i tried
    <br />
\lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)=\\<br />
\lim_{x->\infty}\left (\frac{\frac{1}{\sqrt{x}}\sqrt{x+\sqrt{x+\sqrt{x}}  }-\sqrt{x}}{\frac{1}{\sqrt{x}}} \right)<br />
    but i get 0/0

    ??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,366
    Thanks
    40
    Quote Originally Posted by transgalactic View Post
    i need to solve this limit
    <br />
\lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)<br />
    i tried
    <br />
\lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)=\\<br />
\lim_{x->\infty}\left (\frac{\frac{1}{\sqrt{x}}\sqrt{x+\sqrt{x+\sqrt{x}}  }-\sqrt{x}}{\frac{1}{\sqrt{x}}} \right)<br />
    but i get 0/0

    ??
    First let x = \frac{1}{t^2}, then the limit becomes

    \lim_{t \to 0} \sqrt{\frac{1}{t^2} + \sqrt{\frac{1}{t^2} + \frac{1}{t}}} - \frac{1}{t} = \lim_{t \to 0} \frac{\sqrt{1+t\sqrt{t+1}} - 1}{t}
    Then rationalize

     \lim_{t \to 0} \frac{\sqrt{1+t\sqrt{t+1}} - 1}{t} \cdot \frac{\sqrt{1+t\sqrt{t+1}} + 1}{\sqrt{1+t\sqrt{t+1}} + 1}

    After expanding and canceling you get

     \lim_{t \to 0} \frac{\sqrt{t+1}}{\sqrt{1+t\sqrt{t+1}} + 1} = \frac{1}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,735
    Thanks
    642
    Hello, transgalactic!

    We can solve this head-on ... if we have the stamina.


    \lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)
    Multiply top and bottom by the conjugate . . .

    \frac{\sqrt{x + \sqrt{x + \sqrt{x}}} - \sqrt{x}} {1} \cdot \frac{\sqrt{x+\sqrt{x + \sqrt{x}}} + \sqrt{x}} {\sqrt{x + \sqrt{x + \sqrt{x}}} + \sqrt{x}}

    . . = \;\frac{(x + \sqrt{x+\sqrt{x}}) - x}{\sqrt{x+\sqrt{x+\sqrt{x}}} + \sqrt{x}} \;=\;\frac{\sqrt{x+\sqrt{x}}} {\sqrt{x+\sqrt{x+\sqrt{x}}} + \sqrt{x}}


    Divide top and bottom by \sqrt{x}

    . . \frac{ \dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x}}} {\dfrac{\sqrt{x+\sqrt{x + \sqrt{x}}}}{\sqrt{x}} + \dfrac{\sqrt{x}}{\sqrt{x}}} \;= \; \frac{\sqrt{\dfrac{x}{x} + \dfrac{\sqrt{x}}{x}}} {\sqrt{\dfrac{x}{x} + \dfrac{\sqrt{x+\sqrt{x}}}{x}} + 1 } . =\;\frac{\sqrt{1 + \dfrac{1}{\sqrt{x}}}}   {\sqrt{1 + \sqrt{\dfrac{1}{x} + \dfrac{1}{\sqrt{x^3}}}} + 1 }



    \lim_{x\to\infty}\left(\frac{\sqrt{1 + \dfrac{1}{\sqrt{x}}}}   {\sqrt{1 + \sqrt{\dfrac{1}{x} + \dfrac{1}{\sqrt{x^3}}}} + 1 }\right)  \;=\;\frac{\sqrt{1+0}}{\sqrt{1+\sqrt{0+0}} + 1} \;=\;\frac{1}{1+1} \;=\;\boxed{\frac{1}{2}}



    . . . . . . . \boxed{\begin{array}{c}\text{Kids, don't try this at home.} \\ \text{I'm a trained professional.}\end{array}}


    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. limit of (sqrt(3x+6)-3)/(x-1) | x->1
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 9th 2010, 11:06 AM
  2. Replies: 2
    Last Post: April 21st 2010, 07:27 PM
  3. Limit of sqrt(2*sqrt(2)) sequence
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 4th 2010, 04:06 AM
  4. Replies: 1
    Last Post: April 27th 2009, 06:02 AM
  5. Replies: 11
    Last Post: January 6th 2008, 09:33 AM

Search Tags


/mathhelpforum @mathhelpforum