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Thread: Calculate the limit as n-> infinity

  1. #1
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    Calculate the limit as n-> infinity

    $\displaystyle \lim_{n\rightarrow \infty}\frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1 }$

    How does the n^2/3 distribute?
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  2. #2
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    Re: Calculate the limit as n-> infinity

    $\displaystyle n^{\frac{2}{3}}=\sqrt{n^{\frac{4}{3}}}$
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    Re: Calculate the limit as n-> infinity

    How? Forgive my innocence.
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    Re: Calculate the limit as n-> infinity

    Quote Originally Posted by superduper1 View Post
    How? Forgive my innocence.
    $\displaystyle n^{\frac{2}{3}} * n^{\frac{2}{3}} = n^{\frac{4}{3}} = (n^{\frac{2}{3}})^2$ So Square root that to get $\displaystyle n^{\frac{2}{3}}$
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    Re: Calculate the limit as n-> infinity

    $\displaystyle \lim_{n\rightarrow \infty}\frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1 }$

    $\displaystyle \lim_{n\rightarrow \infty} \frac{\sqrt{n^\frac{7}{3}+2n^\frac{4}{3}}}{2n^2+n+ 1}$

    What should I do next?
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  6. #6
    MHF Contributor MarkFL's Avatar
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    Re: Calculate the limit as n-> infinity

    Basically, if you take a positive expression under a square root, you double the exponent. In general:

    $\displaystyle a^b=\sqrt[n]{a^{bn}}$ where $\displaystyle 0\le a^b$ if $\displaystyle n$ is even and $\displaystyle b$ is odd.
    Last edited by MarkFL; Nov 24th 2012 at 08:50 PM.
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  7. #7
    MHF Contributor MarkFL's Avatar
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    Re: Calculate the limit as n-> infinity

    Quote Originally Posted by superduper1 View Post
    $\displaystyle \lim_{n\rightarrow \infty}\frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1 }$

    $\displaystyle \lim_{n\rightarrow \infty} \frac{\sqrt{n^\frac{7}{3}+2n^\frac{4}{3}}}{2n^2+n+ 1}$

    What should I do next?
    I would put the denominator also under the radical and use the property:

    $\displaystyle \lim_{x\to c}\sqrt{f(x)}=\sqrt{\lim_{x\to c}f(x)}$
    Last edited by MarkFL; Nov 24th 2012 at 08:50 PM.
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  8. #8
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    Re: Calculate the limit as n-> infinity

    solutions says it is 1/2. I keep getting 1/infinity
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    MHF Contributor MarkFL's Avatar
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    Re: Calculate the limit as n-> infinity

    I get zero as well, and so does W|A. Are you certain you have stated the problem correctly? The exponent on n outside the radical in the numerator would have to be 7/2 in order for the limit to be 1/2.
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  10. #10
    MHF Contributor MarkFL's Avatar
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    Re: Calculate the limit as n-> infinity

    Oops, sorry make that 3/2, not 7/2, which makes more sense for a type to have occured.
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    Re: Calculate the limit as n-> infinity

    Quote Originally Posted by superduper1 View Post
    $\displaystyle \lim_{n\rightarrow \infty}\frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1 }$
    How does the n^2/3 distribute?

    $\displaystyle \frac{n^{\frac{2}{3}}(\sqrt{n+2})}{2n^2+n+1}\le \frac{n^{\frac{2}{3}}n^{\frac{1}{2}}(\sqrt{2})}{2n ^2+n+1}$

    Now use comparison.
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