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Math Help - Help in "Limits"...

  1. #1
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    Help in "Limits"...

    Please help in finding the limits of the following..

    1) when limit n -> infinity, find [ sqrt(n^2 + n) - sqrt(n^2 + 1)]
    I know we have to use the rationalizing the denominator but I keep getting this is divergent.

    2) when limit n -> infinity, find (sqrt(n)[ sqrt(n+1) - sqrt(n)]

    3) when limit n -> infinity, find n! / 2^n.
    I know we have to use the ratio test in here, but I am stuck at (n+1)/2.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Vedicmaths View Post
    Please help in finding the limits of the following..

    1) when limit n -> infinity, find [ sqrt(n^2 + n) - sqrt(n^2 + 1)]
    I know we have to use the rationalizing the denominator but I keep getting this is divergent.

    2) when limit n -> infinity, find (sqrt(n)[ sqrt(n+1) - sqrt(n)]

    3) when limit n -> infinity, find n! / 2^n.
    I know we have to use the ratio test in here, but I am stuck at (n+1)/2.
    For the first one multiply by the conjugate to get

    \lim_{n\to\infty}\bigg[\sqrt{n^2+n}-\sqrt{n^2+1}\bigg]\cdot\frac{\sqrt{n^2+n}+\sqrt{n^2+1}}{\sqrt{n^2+n}  +\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{n^2+n-n^2+1}{\sqrt{n^2+n}+\sqrt{n^2+1}}

    Can you go from there?
    For the second one try the same trick

    For the third one

    use the fact taht

    n!\sim{\sqrt{2\pi{n}}n^ne^{-n}}

    to get

    \lim_{n\to\infty}\frac{n!}{2^n}=\lim_{n\to\infty}\  frac{\sqrt{2\pi{n}}n^ne^{-n}}{2^n}=\infty

    But that wouldnt be the Ratio test I dont think?

    If you meant to find the convergence of that series, we just showed by the n-th term test that it is divergent, no need for ratio test
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  3. #3
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    Quote Originally Posted by Vedicmaths View Post
    3) when limit n -> infinity, find n! / 2^n.
    I know we have to use the ratio test in here, but I am stuck at (n+1)/2.
    Instead look at \frac{2^n}{n!}.
    \frac{2^{n+1}}{(n+1)!}\cdot \frac{n!}{2^n} = \frac{2}{n+1} \to 0.
    Thus, \frac{2^n}{n!}\to 0 \implies \frac{n!}{2^n}\to \infty
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    Quote Originally Posted by Vedicmaths View Post
    Please help in finding the limits of the following..

    1) when limit n -> infinity, find [ sqrt(n^2 + n) - sqrt(n^2 + 1)]
    I know we have to use the rationalizing the denominator but I keep getting this is divergent.

    2) when limit n -> infinity, find (sqrt(n)[ sqrt(n+1) - sqrt(n)]

    3) when limit n -> infinity, find n! / 2^n.
    I know we have to use the ratio test in here, but I am stuck at (n+1)/2.
    3) The ratio test requires you to consider \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right|.

    a_n = \frac{n!}{2^n}

    a_{n+1} = \frac{(n+1)!}{2^{n+1}} = \frac{(n+1)n!}{(2) \, 2^n}

    Therefore \left|\frac{a_{n+1}}{a_n}\right| = \frac{n+1}{2}.

    Therefore  \lim_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n}\right| = \lim_{n \rightarrow \infty} \frac{n+1}{2} > 1.

    Therefore .......
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  5. #5
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    Dear Mathstud 28!

    Thanks for the prompt response. Sir, for the second one, I myself got until (n-1) / [ sqrt(n^2 + n) + sqrt(n^2 + 1)] but then after that I am stuck and this is the same case with the second one too.
    Do we have to split that in two forms like (a-b)/c is a/c - b/c and then try solving it??

    Thanks again!
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Vedicmaths View Post
    Please help in finding the limits of the following..

    1) when limit n -> infinity, find [ sqrt(n^2 + n) - sqrt(n^2 + 1)]
    I know we have to use the rationalizing the denominator but I keep getting this is divergent.

    2) when limit n -> infinity, find (sqrt(n)[ sqrt(n+1) - sqrt(n)]

    3) when limit n -> infinity, find n! / 2^n.
    I know we have to use the ratio test in here, but I am stuck at (n+1)/2.
    Quote Originally Posted by Vedicmaths View Post
    Dear Mathstud 28!

    Thanks for the prompt response. Sir, for the second one, I myself got until (n-1) / [ sqrt(n^2 + n) + sqrt(n^2 + 1)] but then after that I am stuck and this is the same case with the second one too.
    Do we have to split that in two forms like (a-b)/c is a/c - b/c and then try solving it??

    Thanks again!
    Consider the fact that as n\to\infty \sqrt{n^2+n}+\sqrt{n^2+1}\to{2n}
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    Quote Originally Posted by Vedicmaths View Post
    [snip]
    2) when limit n -> infinity, find (sqrt(n)[ sqrt(n+1) - sqrt(n)]
    [snip]
    \frac{\sqrt{n} \, (\sqrt{n+1} - \sqrt{n})}{1} \times \frac{(\sqrt{n+1} + \sqrt{n})}{(\sqrt{n+1} + \sqrt{n})} = \frac{\sqrt{n}}{\sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1} ......
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  8. #8
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    Quote Originally Posted by Vedicmaths View Post
    Dear Mathstud 28!

    Thanks for the prompt response. Sir, for the second one, I myself got until (n-1) / [ sqrt(n^2 + n) + sqrt(n^2 + 1)] but then after that I am stuck and this is the same case with the second one too.
    Do we have to split that in two forms like (a-b)/c is a/c - b/c and then try solving it??

    Thanks again!
    \lim_{n\to\infty}\bigg[\sqrt{n^2+n}-\sqrt{n^2+1}\bigg]\cdot\frac{\sqrt{n^2+n}+\sqrt{n^2+1}}{\sqrt{n^2+n}  +\sqrt{n^2+1}}=\lim_{n\to\infty}\frac{n^2+n-n^2+1}{\sqrt{n^2+n}+\sqrt{n^2+1}}

    = \lim_{n\to\infty}\frac{n+1}{\sqrt{n^2+n}+\sqrt{n^2  +1}} = \lim_{n\to\infty}\frac{1+\frac{1}{n}}{\sqrt{1 + \frac{1}{n}} +\sqrt{1 + \frac{1}{n^2}}}
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  9. #9
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    Thanks a lot guys!

    That helps a lot!
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