1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook