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Thread: limit and dividing by n^(3/4) CORRECTION

  1. #1
    TS1
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    limit and dividing by n^(3/4) CORRECTION

    this is a simple question but i cant seem to get it.
    I have to find

    $\displaystyle
    [\lim_{n \to \infty} (n^{\frac{3}{4}})((n+1)^{\frac{1}{4}}- (n-1)^{\frac{1}{4}})
    $

    i have got to

    $\displaystyle
    [\lim_{n \to \infty} \frac{(n^{\frac{3}{4}})(2)}{(n+1)^{\frac{3}{4}}+(n +1)^{\frac{2}{4}}(n-1)^{\frac{1}{4}}+(n+1)^{\frac{1}{4}}(n-1)^{\frac{2}{4}}+(n-1)^{\frac{3}{4}}}
    $


    and know that i have to divide by n^(3/4) to get the answer 1/2 as n tend to infinity.
    I keep making mistakes when dividing by this, can anyone help step by step?
    Thanks
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  2. #2
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    Quote Originally Posted by TS1 View Post
    this is a simple question but i cant seem to get it.
    I have to find

    $\displaystyle
    [\lim_{n \to \infty} (n^{\frac{3}{4}})((n+1)^{\frac{1}{4}}- (n-1)^{\frac{1}{4}})
    $

    i have got to

    $\displaystyle
    [\lim_{n \to \infty} \frac{(n^{\frac{3}{4}})(2)}{(n+1)^{\frac{3}{4}}+(n +1)^{\frac{2}{4}}(n-1)^{\frac{1}{4}}+(n+1)^{\frac{1}{4}}(n-1)^{\frac{2}{4}}+(n-1)^{\frac{3}{4}}}
    $


    and know that i have to divide by n^(3/4) to get the answer 1/2 as n tend to infinity.
    I keep making mistakes when dividing by this, can anyone help step by step?
    Thanks
    Dear TS1,

    Can you tell me the answers for,

    $\displaystyle \frac{(n+1)^{\frac{3}{4}}}{n^{\frac{3}{4}}}=$ ?

    $\displaystyle \frac{(n+1)^{\frac{2}{4}}(n-1)^{\frac{1}{4}}}{n^{\frac{3}{4}}}=$ ?
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  3. #3
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    To the given problem, multiply and divide by
    [(n+1)^1/4 + (n-1)^1/4][(n+1)^1/2 + (n-1)^1/2]
    and simplify.
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  4. #4
    TS1
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    $\displaystyle
    \frac{(n+1)^{\frac{3}{4}}}{n^{\frac{3}{4}}}=
    $

    (1+1/n)^(3/4)?


    i cant do these though

    $\displaystyle
    \frac{(n+1)^{\frac{2}{4}}(n-1)^{\frac{1}{4}}}{n^{\frac{3}{4}}}=
    $

    $\displaystyle
    \frac{(n+1)^{\frac{1}{4}}(n-1)^{\frac{2}{4}}}{n^{\frac{3}{4}}}=
    $
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  5. #5
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    Your first simplification is correct.
    Now as n tends to infinity 1/n tends to zero.

    Similarly you can simplify the other two.
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  6. #6
    TS1
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    how do i do that when the powers are different?
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  7. #7
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    Hello, TS1!

    You are that close . . .


    I have got to:
    $\displaystyle \lim_{n \to \infty} \frac{(n^{\frac{3}{4}})(2)}{(n+1)^{\frac{3}{4}}+(n +1)^{\frac{2}{4}}(n-1)^{\frac{1}{4}}+(n+1)^{\frac{1}{4}}(n-1)^{\frac{2}{4}}+(n-1)^{\frac{3}{4}}}$

    I know that I have to divide by $\displaystyle n^{\frac{3}{4}}$ to get the answer $\displaystyle \tfrac{1}{2}.$ . . Right!

    Divide top and bottom by $\displaystyle n^{\frac{3}{4}}$

    $\displaystyle \frac{2}{\dfrac{(n+1)^{\frac{3}{4}}}{n^{\frac{3}{4 }}} + \dfrac{(n+1)^{\frac{2}{4}}}{n^{\frac{2}{4}}} \dfrac{(n-1)^{\frac{1}{4}}}{n^{\frac{1}{4}}} + \dfrac{(n+1)^{\frac{1}{4}}}{n^{\frac{1}{4}}}\dfrac {(n-1)^{\frac{2}{4}}}{n^{\frac{2}{4}}} + \dfrac{(n-1)^{\frac{3}{4}}}{n^{\frac{3}{4}}}}$


    . . $\displaystyle =\; \frac{2} {\left(\dfrac{n+1}{n}\right)^{\frac{3}{4}} + \left(\dfrac{n+1}{n}\right)^{\frac{2}{4}} \left(\dfrac{n-1}{n}\right)^{\frac{1}{4}} + \left(\dfrac{n+1}{n}\right)^{\frac{1}{4}}
    \left(\dfrac{n-1}{n}\right)^{\frac{2}{4}} + \left(\dfrac{n-1}{n}\right)^{\frac{3}{4}}}$


    . . $\displaystyle =\; \frac{2} {\left(1 + \dfrac{1}{n}\right)^{\frac{3}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{2}{4}} \left(1 - \dfrac{1}{n}\right)^{\frac{1}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{1}{4}}\left(1 - \dfrac{1}{n}\right)^{\frac{2}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{3}{4}}}$



    Then: . $\displaystyle \lim_{n\to\infty}\left[ \frac{2} {\left(1 + \dfrac{1}{n}\right)^{\frac{3}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{2}{4}} \left(1 - \dfrac{1}{n}\right)^{\frac{1}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{1}{4}}\left(1 - \dfrac{1}{n}\right)^{\frac{2}{4}} + \left(1 + \dfrac{1}{n}\right)^{\frac{3}{4}}}\right]$


    . . . . $\displaystyle =\;\; \frac{2} {(1 + 0)^{\frac{3}{4}} + (1 + 0)^{\frac{2}{4}} (1 - 0)^{\frac{1}{4}} + (1 + 0)^{\frac{1}{4}}(1 - 0)^{\frac{2}{4}} + (1 + 0)^{\frac{3}{4}}}$


    . . . . $\displaystyle =\;\;\frac{2}{1+ 1\!\cdot\!1 + 1\!\cdot\!1 + 1 } \;\;=\;\;\frac{2}{4} \;\;=\;\;\frac{1}{2}$

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