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Math Help - strange... limit

  1. #1
    Member disclaimer's Avatar
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    strange... limit

    Hello;

    \lim_{n\rightarrow\infty}\sqrt{n^{10}-2n^2+2}

    To me, it's \infty, as I think we can write:

    \lim_{n\rightarrow\infty}\sqrt{n^{10}-2n^2+2}= \lim_{n\rightarrow\infty}\sqrt{n^{10}\left(1-\frac{2}{n^8}+\frac{2}{n^{10}}\right)}= \lim_{n\rightarrow\infty}\left(n^5\sqrt{1-\frac{2}{n^8}+\frac{2}{n^{10}}}\right)= \infty\cdot1=\infty

    But the answer in my textbook is 1. Am I missing something here?

    Help appreciated.
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  2. #2
    Super Member wingless's Avatar
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    \lim_{n\rightarrow\infty}\sqrt{n^{10}-2n^2+2} = \infty

    Are you sure that it's \lim_{n\rightarrow\infty} instead of \lim_{n\rightarrow 1} or something? Because the limit approaching to infinity is infinity (as you found), not 1.
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  3. #3
    Flow Master
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    Quote Originally Posted by disclaimer View Post
    Hello;

    \lim_{n\rightarrow\infty}\sqrt{n^{10}-2n^2+2}

    To me, it's \infty, as I think we can write:

    \lim_{n\rightarrow\infty}\sqrt{n^{10}-2n^2+2}= \lim_{n\rightarrow\infty}\sqrt{n^{10}\left(1-\frac{2}{n^8}+\frac{2}{n^{10}}\right)}= \lim_{n\rightarrow\infty}\left(n^5\sqrt{1-\frac{2}{n^8}+\frac{2}{n^{10}}}\right)= \infty\cdot1=\infty

    But the answer in my textbook is 1. Am I missing something here? Mr F says: No. But your book probably is ..... the print for the continuation of the expression that you're taking the limit of.

    Help appreciated.
    The question was probably meant to be something like

    \lim_{n\rightarrow\infty} (\sqrt{n^{10}-2n^2+2} - \sqrt{n^{10} + \text{polynomial stuff of degree less than 10}}\, ).
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  4. #4
    Member disclaimer's Avatar
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    Well, in the book they give just u_n=\sqrt{n^{10}-2n^2+2} and the problem is to find the limit of the sequence having such a general formula.

    Thanks for taking the time to look into that.
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  5. #5
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    Could it be u_n  = \sqrt[n]{{n^{10}  - 2n^2 + 2}}?
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