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Math Help - Limits of sequences

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    Limits of sequences

    What are the limits of the follwing sequences as n tends towards infinity?

    (i) \sqrt {{n}^{2}+1000}-n

    (ii) \left(  \left( n! \right) ^{{n}^{-2}} \right) ^{-1}

    (iii) {\frac {{n}^{100}}{{ 1.001}^{n}}}

    (iv) {\frac {n \left( 2\,n+3 \right)  \left( 3\,n+4 \right) +2}{4\, \left( <br />
n \left( 6\,n+7 \right)  \right) ^{2}+5\,n \left( 6\,n+7 \right) +6}}

    I had a go at (iv) and got an answer 1/4....

    Im not really sure how to do the others, they are part of a past exam paper and there is no mark scheme, so when i get to an answer im not sure if it is right or not!
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    I think (i) is 0 and (ii) is 1.... but i am only coming to these answers intuitively.... is there a way to verify them?
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    Quote Originally Posted by Mathman87 View Post
    I think (i) is 0 and (ii) is 1.... but i am only coming to these answers intuitively.... is there a way to verify them?
    (i) is 0
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    (ii) re-write the equation x^{-1}=\frac{1}{x}
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    (iv) \frac{6n^3}{144n^4}=\frac{1}{24n}
    So the limit is 0
    Last edited by dwsmith; May 11th 2010 at 08:41 PM. Reason: I forgot a n term on the bottom yielding the wrong limit
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    Quote Originally Posted by Mathman87 View Post
    I think (i) is 0 and (ii) is 1.... but i am only coming to these answers intuitively.... is there a way to verify them?

    Using intuition is a first step to solving problems , then you may solve them by following your intuition .

    For example , for question (i) , intuitively it is zero , we can do something to prove it zero . We know  \frac{1}{\infty} is zero so can we do something to rewrite the limit in this form ?

     \sqrt{n^2 + 1000} - n

     = (\sqrt{n^2 + 1000} - n  ) \cdot \frac{\sqrt{n^2 + 1000} + n }{\sqrt{n^2 + 1000} + n }

     = \frac{(n^2+1000)-n^2 }{ \sqrt{n^2 + 1000} + n } .
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    (iii) is 0 believe it or not.

    \frac {n^{100}}{1.001^n} = \Big(\frac {n^{\frac{100}{n}}}{1.001}\Big)^n

    \frac{100}{n}\to 0 \implies \Big(\frac {n^{\frac{100}{n}}}{1.001}\Big)^n \to 0.
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    Quote Originally Posted by Anonymous1 View Post
    (iii) is 0 believe it or not.
    I believe it! Maple told me so.
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    Misread question
    Last edited by HallsofIvy; May 12th 2010 at 03:02 AM.
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  10. #10
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    Quote Originally Posted by Mathman87 View Post
    What are the limits of the follwing sequences as n tends towards infinity?

    (i) \sqrt {{n}^{2}+1000}-n
    Multiply numerator and denominator by \sqrt{n^2+ 1000}+ n:
    (\sqrt{n^2+ 1000}- n)\frac{\sqrt{n^2+ 1000}+ n}{\sqrt{n^2+ 1000}+ n}= \frac{n^2+ 1000- n^2}{\sqrt{n^2+ 1000}+ n}= \frac{1000}{\sqrt{n^2+ 1000}+ n}.

    (ii) \left(  \left( n! \right) ^{{n}^{-2}} \right) ^{-1}
    This is the same as \frac{1}{(n!)^{n^2}}. The numerator is 1 while the denominator gets bigger and bigger without bound.

    (iii) {\frac {{n}^{100}}{{ 1.001}^{n}}}
    Probably the simplest thing to do is use L'Hopital's rule- 100 times!

    (iv) {\frac {n \left( 2\,n+3 \right)  \left( 3\,n+4 \right) +2}{4\, \left( <br />
n \left( 6\,n+7 \right)  \right) ^{2}+5\,n \left( 6\,n+7 \right) +6}}

    I had a go at (iv) and got an answer 1/4....
    If you were to multiply out the numerator, the leading term would be 6n^3. Doing the same in the denominator the leading term would be 144n^4. Divide both numerator and denominator by n^4.

    Im not really sure how to do the others, they are part of a past exam paper and there is no mark scheme, so when i get to an answer im not sure if it is right or not!
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  11. #11
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Probably the simplest thing to do is use L'Hopital's rule- 100 times!
    I'm vaguely recalling some theorem from R-Analysis you could quote here...

    Something about exponential growth "getting there faster."
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