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Limits using basic analysis theorems and logic?

  1. Aug 20, 2005 #1
    Hi,

    I need help again. How can I show that

    [tex]\sqrt{n^2+2n}-n \rightarrow 1[/tex]​

    for [tex]n\rightarrow\infty[/tex] using basic analysis theorems and logic? Any ideas?
     
  2. jcsd
  3. Aug 20, 2005 #2

    Hurkyl

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    I bet if you checked your book, you'd find an example of a clever way of dealing with limits involving differences like these...
     
  4. Aug 20, 2005 #3
    try doing something to it that involves its conjugate...
     
  5. Aug 20, 2005 #4

    CarlB

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    Rewrite [tex]\sqrt{n^2+2n}-n [/tex] as


    [tex][\sqrt{(n+1)^2 - 1^2} -(n+1)] + 1[/tex]

    Thus you need to show that

    [tex]\sqrt{(n+1)^2 - 1^2} -(n+1) \rightarrow 0[/tex]

    as n to infinity. But if you look at the way it is written, you will see that it looks suspiciously like the difference between the length of two sides of a right triangle as one of the points on the triangle goes to infinity.

    Carl
     
  6. Aug 21, 2005 #5
    Thanks guys. Hurkyl was right; there actually was an example dealing with this in the back of the book. I can't believe I missed it.
     
  7. Aug 21, 2005 #6

    quasar987

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    What is the solution?
     
  8. Aug 21, 2005 #7
    We want to show that [tex]\sqrt{n^2+2n} - (n+1) \rightarrow 0[/tex], so we rewrite the expression:


    [tex]\sqrt{n^2+2n} - (n+1) = \frac{\left(\sqrt{n^2+2n} - (n+1)\right)\left(\sqrt{n^2+2n} + (n+1)\right)}{\sqrt{n^2+2n} + (n+1)} = \frac{-1}{\sqrt{n^2+2n} + (n+1)}[/tex]


    It is obvious that the final expression tends to 0 as [tex]n\rightarrow\infty[/tex].
     
  9. Aug 22, 2005 #8
    I solved it thus:

    (n^2 + 2n)^1/2 = (n^2)^1/2 * (1+2/n)^1/2


    therefore this = n(1+2/n)^1/2

    now expand the bracket using taylor's theorem to get

    (1+2/n)^1/2= 1+(1/n)-(1/n^2)+...


    therefore multiplying by n in the above you have n+1-(1/n) as n tends to infinity the (1/n) becomes irrelevent and we are left with n+1. Just subtract the n from the original expression and you obtain the desired result, 1.

    Take care, Joe
     
  10. Aug 23, 2005 #9

    HallsofIvy

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    Gosh, that looks complicated!

    I would have thought using gDogg's suggestion and writing
    [tex]\sqrt{n^2+2n}-n= \frac{(\sqrt{n^2+2n}-n)(\sqrt{n^2+2n}+n)}{\sqrt{n^2+2n}+n}[/tex]
    [tex]= \frac{2n}{\sqrt{n^2+ 2n}+n}[/tex]

    would be much simpler.
     
  11. Aug 23, 2005 #10

    Hurkyl

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    Yes... he could have used the binomial theorem!

    [tex]
    (n^2 + 2n)^{1/2} = (n^2)^{1/2} + \frac{1/2}{1} (n^2)^{-1/2} (2n) + \frac{(1/2) * (-1/2)}{1 * 2} (n^2)^{-3/2} (2n)^2 + \cdots
    = n + 1 + O\left(\frac{1}{n}\right)
    [/tex]

    (since n² > 2n when n grows large)
    (Hrm, did he make a mistake in the third term, or did I?)
     
    Last edited: Aug 23, 2005
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