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Thread: How to calculate this limit?

  1. #1
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    How to calculate this limit?

    $\displaystyle \lim_{x \to \infty_+}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$

    Thank you very much!
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  2. #2
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    Re: How to calculate this limit?

    Hello, gotmejerry!

    Nice job with the Latex!


    $\displaystyle \lim_{x \to \infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$

    Multiply numerator and denominator 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 =\;\dfrac{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 numerator and denominator by $\displaystyle \sqrt{x}\!:$

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

    . . $\displaystyle =\; \dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}} {\sqrt{1 + \dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x^2}}} + 1} \;=\;\dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}} {\sqrt{1 + \sqrt{\dfrac{x}{x^2} + \dfrac{\sqrt{x}}{x^2}}}+ 1}\;=\;\dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}} {\sqrt{1 + \sqrt{\dfrac{1}{x} + \dfrac{1}{x^{\frac{3}{2}}}}}+ 1} $


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

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  3. #3
    Junior Member
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    Re: How to calculate this limit?

    Thank you very much! Your LATEX is very good too
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