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Math Help - Wrong limit value, why?

  1. #1
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    Wrong limit value, why?

    Hello, first of all i might excuse myself for my poor english, i'm Italian and my native language is far from being "english-friendly";

    down to the question itself i had a test at university where i was called to solve this limit:

    i might also want to specify that we could not use De L'Hopital or other "advanced tools" as we're studying "Sequences' limits" and not "Functions' limits", so we have to basically reduce this to a "collection of known limits" to be able to specify the value it assumes when n->+infinity.

    Well, my solution said that the Sequence was convergent to 0, while Wolfram Alpha (as you can see from the image) says that it converges to 1...
    How would you guys have approached the solution of said limit?

    Thanks in advance,
    SkyWolf.
    Attached Thumbnails Attached Thumbnails Wrong limit value, why?-2qkmsrq.png  
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    If you are allowed to use:

    \displaystyle\lim_{n \to{+}\infty}{\dfrac{\log (1+a_n)}{a_n}}=1\quad (a_n\to 0)

    then, you can substitute the denominator by:

    a_n=\dfrac{n}{n^3+1}

    and you can easily find the limit.
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  3. #3
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    Quote Originally Posted by FernandoRevilla View Post
    .. and you can easily find the limit.
    Yeah... You are perfectly right; darn it ... One just needs the intuition to replace things with the right known limit to solve an exercise with the speed of light...

    Muchas gracias,
    SkyWolf.
    Last edited by mr fantastic; August 22nd 2011 at 04:54 AM.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by SkyWolf View Post
    Muchas gracias,

    Benvenuti.
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by SkyWolf View Post
    Hello, first of all i might excuse myself for my poor english, i'm Italian and my native language is far from being "english-friendly";

    down to the question itself i had a test at university where i was called to solve this limit:

    i might also want to specify that we could not use De L'Hopital or other "advanced tools" as we're studying "Sequences' limits" and not "Functions' limits", so we have to basically reduce this to a "collection of known limits" to be able to specify the value it assumes when n->+infinity.

    Well, my solution said that the Sequence was convergent to 0, while Wolfram Alpha (as you can see from the image) says that it converges to 1...
    How would you guys have approached the solution of said limit?

    Thanks in advance,
    SkyWolf.
    First You transform the expression as follows...

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

    ... so that the problem is to find the limit of the denominator. Then You take into account that is ...

    \displaystyle \ln (1+\frac{n}{1+n^{3}})= \frac{n}{1+n^{3}} - \frac{n^{2}}{2\ (1+n^{3})^{2}} + ... (2)

    ... and...

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

    ... so that is...

    \displaystyle \lim_{n \rightarrow \infty} \ln (1+\frac{n}{1+n^{3}})\ \{\sqrt{n^{4}+1}+\sqrt{n^{4}-1}\}} = 2 (4)

    ... and the 'goal' is met!...

    Kind regards

    \chi \sigma

    P.S. I'm Italian too and also my English is a little 'imperfect' ... never mind!... we have allways the opportunity to improve ourselves!...
    Last edited by chisigma; February 27th 2011 at 10:18 AM.
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