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Thread: sqrt limit question..

  1. #1
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    sqrt limit question..

    i need to solve this limit
    $\displaystyle
    \lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)
    $
    i tried
    $\displaystyle
    \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)
    $
    but i get 0/0

    ??
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    i need to solve this limit
    $\displaystyle
    \lim_{x->\infty}\left ( \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)
    $
    i tried
    $\displaystyle
    \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)
    $
    but i get 0/0

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

    $\displaystyle \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

    $\displaystyle \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

    $\displaystyle \lim_{t \to 0} \frac{\sqrt{t+1}}{\sqrt{1+t\sqrt{t+1}} + 1} = \frac{1}{2}$
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  3. #3
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    Hello, transgalactic!

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


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

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

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


    Divide top and bottom by $\displaystyle \sqrt{x}$

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



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



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


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