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Homework Help: Proof of Limit

  1. Jan 23, 2005 #1
    Hello all

    I need help with the following problems:

    Prove that [tex] \lim_{x\rightarrow \infty} (\sqrt{n+1} - \sqrt n)(\sqrt {n+ \frac {1}{2}}) = \frac {1}{2} [/tex]

    I know that [tex] \lim_{x\rightarrow \infty} (\sqrt{n+1} - \sqrt n) = 0 [/tex]. Then why wouldn't the limit be 0?

    Also another question (I posted this in another thread, but it died :tongue2:)

    [tex] a_n = \sqrt {n+1} - \sqrt n [/tex] find three numbers [tex] N_1 , N_2, N_3 [/tex] such that

    [tex] a_n = \sqrt {n+1} - \sqrt n < \frac {1}{10} [/tex] for every [tex] n > N_1 [/tex]
    [tex] a_n = \sqrt {n+1} - \sqrt n < \frac {1}{100} [/tex] for every [tex] n > N_2 [/tex]
    [tex] a_n = \sqrt {n+1} - \sqrt n < \frac {1}{1000} [/tex] for every [tex] n > N_3 [/tex]
  2. jcsd
  3. Jan 23, 2005 #2


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    Reexpress [tex]\sqrt{n + a}[/tex] as [tex]\sqrt{n}\sqrt{1 + \frac{a}{n}[/tex] and use the general binomial expansion to simplify the series. You only need the first order approximation.

    That is,

    [tex]\sqrt{1 + \frac{a}{n}} \approx \left(1 + (\frac{1}{2})(\frac{a}{n})\right)[/tex]

    For the first part the a-values are 1 and [itex]\frac{1}{2}[/itex]

    Simplify the algebra, and you'll find the limit.

    For the second part use the simplified expression found in the first part for the expression. You should be able to find a simple inequality in each case and solve accordingly. Post if you need further hints.
    Last edited: Jan 23, 2005
  4. Jan 23, 2005 #3
    Because sqrt(n + 1/2) isn't bounded.

    Surely 1/n tends to 0 as n tends to infinity. But what's the limit of n * 1/n?
  5. Jan 23, 2005 #4
    ok so for the first part, the simplified expression I got was:

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

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

    and this equals [tex] \frac {1}{2} [/tex] ?

    hmm.. it looks like i went in circles
    Last edited: Jan 23, 2005
  6. Jan 23, 2005 #5


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    Have you tried using the binomial expansion and simplifying as I advised ?
  7. Jan 23, 2005 #6
    do you mean [tex] \sqrt n (\sqrt{1 + \frac {1}{n}} - \sqrt n)\sqrt n ( \sqrt{1+ \frac {1}{2n}))^n [/tex] ?

    NOTE: This is in the beginning of the book. I haven't learned about series expansions yet

    Last edited: Jan 23, 2005
  8. Jan 23, 2005 #7


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    I think there's a mistake here. I think it should be:

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

    I rewrite as:
    [tex] \frac{(\sqrt{1 + \frac {1}{n}} - 1) ( \sqrt{1+ \frac {1}{2n}})}{\frac{1}{n}}[/tex]

    Substitue x=1/n. So x->0 for the limit.

    Then you can use L'Hopital's rule (since numerator and denominator go to zero), if you've covered that.
  9. Jan 23, 2005 #8
    thanks for the help. but how would you use the binomail expansion? Thats whats confusing me

  10. Jan 23, 2005 #9


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    Have a look at my previous post. Do you know l'Hopital's rule? I personally find this easier, as I tend to make mistakes when substituting the binomial expansion.

    But you can also use binomial expansion:

    [tex]\sqrt {1+x} = {(1+x)}^{(1/2)}= 1 + (1/2)x + [(1/2)((1/2) - 1)/2!]x^2 + .....[/tex]

    Under the first square root x=1/n, under the second x=1/(2n).
  11. Jan 23, 2005 #10
    Thanks a bunch
  12. Jan 23, 2005 #11
    ok so now that i have the binomial expansion, how would i simplify it?
    Would I use [tex]\sqrt{1 + \frac{a}{n}} \approx \left(1 + (\frac{1}{2})(\frac{a}{n})\right)[/tex] which when we substitute equals:

    [tex] (1 + \frac{1}{2n})(1 + \frac{1}{4n}) = (1 + \frac{1}{4n} + \frac{1}{2n} + \frac{1}{8n^2})(\sqrt n) [/tex]

    Last edited: Jan 23, 2005
  13. Jan 23, 2005 #12


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    If you don't know the "binomial expansion", you shouldn't be trying to use it for this problem!

    [tex] \lim_{x\rightarrow \infty} (\sqrt{n+1} - \sqrt n)(\sqrt {n+ \frac {1}{2}}) = \frac {1}{2} [/tex]

    There's nothing wrong with going ahead and multiplying those:

    [tex]\sqrt{n^2+ \frac{3}{2}n+ \frac{1}{2}}-\sqrt{n^2+ \frac{1}{2}n}[/tex]

    Now do the usual "trick" of rationalizing the numerator: multiply both numerator and denominator (which is 1) by
    [tex]\sqrt{n^2+ \frac{3}{2}n+ \frac{1}{2}}+\sqrt{n^2+ \frac{1}{2}n}[/tex]
    That gives
    [tex]\frac{n+ \frac{1}{2}}{\sqrt{n^2+ \frac{3}{2}n+ \frac{1}{2}}+\sqrt{n^2+ \frac{1}{2}n}}[/tex]

    Finally divide both numerator and denominator by n (which becomes n2 inside the square roots) and take the limit as n goes to infinity.

    For the second question, just go ahead and solve the equations!

    Find N1 such that [tex] a_n = \sqrt {n+1} - \sqrt n < \frac {1}{10} [/tex]
    for all n> N1.

    Well, to pass from "<1/10" to ">1/10" it has to be "= 1/10". When is
    [tex] a_n = \sqrt {n+1} - \sqrt n = \frac {1}{10} ?[/tex]

    Write this as [itex]\sqrt{n+1}= \sqrt{n}+ \frac{1}{10}[/itex] and square both sides:

    [itex]n+ 1= n+ \frac{2\sqrt{n}}{10}+ \frac{1}{100}[/itex].
    Rewrite as [itex]\frac{2\sqrt{n}}{10}= 1- \frac{1}{100}= \frac{99}{100}[/itex] and square again: [itex]\frac{4n}{100}= \frac{9801}{10000}[/itex] so
    [itex]n= \frac{9801}{400}= 24.5[/itex] (approx.). Since n has to be an integer, n must be larger than 24: N1= 24.

    Do the same for the others.
    Last edited by a moderator: Jan 23, 2005
  14. Jan 23, 2005 #13
    thank you for your very helpful response HallsofIvy. Also I want to thank everyone that took the time to help me with this problem.
  15. Jan 23, 2005 #14
    just had a quick question"

    for [tex]\frac{n+ \frac{1}{2}}{\sqrt{n^2+ \frac{3}{2}n+ \frac{1}{2}}+\sqrt{n^2+ \frac{1}{2}n}}[/tex] how do we get the denominator after we divide by n?

    EDIT: Nevermind I got it. You divide by n^2 because of the square root sign
    Last edited: Jan 23, 2005
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