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Math Help - Limits question

  1. #1
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    Limits question



    Find the limit or show that no limit exists. I didn't really understand the lecture but I think the limit doesn't exist here. Can anyone help? Thanks.
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    Quote Originally Posted by Thomas154321 View Post


    Find the limit or show that no limit exists. I didn't really understand the lecture but I think the limit doesn't exist here. Can anyone help? Thanks.
    \lim_{x \to \infty} \frac{n^2 + log(n)}{\sqrt{2n^3 - 1}}

    First note that
    \lim_{n \to \infty} \frac{log(n)}{\sqrt{2n^3 - 1} }= 0 <-- You show this.
    so we don't have to worry about that term.

    Thus
    \lim_{x \to \infty} \frac{n^2 + log(n)}{\sqrt{2n^3 - 1}} = \lim_{x \to \infty} \frac{n^2}{\sqrt{2n^3 - 1}}

    = \lim_{x \to \infty} \frac{n^2}{\sqrt{2}n^{3/2}} = \frac{1}{\sqrt{2}} \lim_{x \to \infty} n^{1/2} \to \infty

    -Dan
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    Thanks for the fast reply!

    Just wondering why you can just ignore the 1 in the denominator in the last line.

    This was another one:



    I think the answer is -2 but not sure, can you check please? Thanks so much!
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Thomas154321 View Post
    Thanks for the fast reply!

    Just wondering why you can just ignore the 1 in the denominator in the last line.
    Think of it this (un-math-proof-like) way: As n gets very very large, how does 2n^2 compare to that measly 1?

    I'll leave it to you to formalize this statement.

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Thomas154321 View Post

    \lim_{n \to \infty}n^2 \left ( \frac{1}{n + 1} - \frac{1}{n - 1} \right ) = \lim_{n \to \infty} n^2 \cdot \frac{-2}{(n + 1)(n - 1)}

    = \lim_{n \to \infty} \frac{-2n^2}{n^2 - 1}

    = \lim_{n \to \infty} \frac{-2n^2}{n^2 - 1} \cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}}

    = \lim_{n \to \infty} \frac{-2}{1 - \frac{1}{n^2}} = -2

    -Dan
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