Results 1 to 3 of 3

Math Help - Limit involving square roots

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    22

    Limit involving square roots

    Hi there,

    I am trying to see why this function goes to zero at the limit of large n (for constant a):
    \frac{2}{a}(\sqrt{n}\sqrt{n+a}-2n-a)

    With some factoring out, I've arrived at this:
    \frac{2n}{a}(\sqrt{1+\frac{a}{n}}-1-\frac{a}{2n})

    Where the first (root) term in the bracket should go to 1, and the last to 0, thus the bracket should approach zero in the limit. However, for some reason I don't feel quite comfortable with my method and reasoning. Don't I have to worry about the rate the two terms of the product approach their respective limits? Is there a better way to express the function to make this nuance clear?

    Thanks,

    Chris
    Last edited by entropyslave; June 24th 2012 at 06:38 AM. Reason: Clarification
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,412
    Thanks
    1328

    Re: Limit involving square roots

    Quote Originally Posted by entropyslave View Post
    Hi there,

    I am trying to see why this function goes to zero at the limit of large n (for constant a):
    \frac{2}{a}(\sqrt{n}\sqrt{n+a}-2n-a)

    With some factoring out, I've arrived at this:
    \frac{2n}{a}(\sqrt{1+\frac{a}{n}}-1-\frac{a}{2n})

    Where the first (root) term in the bracket should go to 1, and the last to 0, thus the bracket should approach zero in the limit. However, for some reason I don't feel quite comfortable with my method and reasoning. Don't I have to worry about the rate the two terms of the product approach their respective limits? Is there a better way to express the function to make this nuance clear?
    You should not feel comfortable! Yes, the bracket approaches 0 but then you are multiplying it by a term that goes to infinity- you have an "indeterminate" of the form 0(\infty).
    Instead do what you typically do with problems involving square roots- multiply both numerator and denominator by the "conjugate":
    (\sqrt{n}\sqrt{n+1}- 2n- a)\frac{\sqrt{n}\sqrt{n+1}+ 2n+ a}{\sqrt{n}\sqrt{n+1}+ 2n+ a}= \frac{n(n+1)- (4n^2- 4an+ a^2)}{\sqrt{n}\sqrt{n+1}+ 2n+ a}

    Thanks,

    Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jul 2010
    Posts
    22

    Re: Limit involving square roots

    Quote Originally Posted by HallsofIvy View Post
    You should not feel comfortable! Yes, the bracket approaches 0 but then you are multiplying it by a term that goes to infinity- you have an "indeterminate" of the form 0(\infty).
    Instead do what you typically do with problems involving square roots- multiply both numerator and denominator by the "conjugate":
    (\sqrt{n}\sqrt{n+1}- 2n- a)\frac{\sqrt{n}\sqrt{n+1}+ 2n+ a}{\sqrt{n}\sqrt{n+1}+ 2n+ a}= \frac{n(n+1)- (4n^2- 4an+ a^2)}{\sqrt{n}\sqrt{n+1}+ 2n+ a}
    I see, thanks for making this clearer to me!
    So there was a typo in my original formula - it should of read (with the 2 as part of the first term):
    \frac{1}{a}(2\sqrt{n}\sqrt{n+a}-2n-a)
    So now following your method, I get:
    \frac{1}{a}[2\sqrt{n}\sqrt{n+a}-(2n+a)]\frac{2\sqrt{n}\sqrt{n+a}+(2n+a)}{2\sqrt{n}\sqrt{n  +a}+2n+a}=\frac{1}{a}\frac{4n(n+a)-(4n^2+4an+a^2)}{2\sqrt{n}\sqrt{n+a}+(2n+a)}=\frac{-a}{2\sqrt{n}\sqrt{n+a}+2n+a}

    And now this already clearly goes to zero, but I think it is even clearer if I apply a similar factorisation as my original post:
    =\frac{-a}{2n(1+\sqrt{1+\frac{a}{n}}+\frac{a}{2n})}
    Which for large n looks like -\frac{a}{4n}

    Thanks for your help!
    Last edited by entropyslave; June 24th 2012 at 08:15 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: September 7th 2011, 07:13 AM
  2. Prove an Equation involving Square Roots
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 11th 2011, 12:04 PM
  3. Inequality involving square roots
    Posted in the Calculus Forum
    Replies: 7
    Last Post: June 28th 2010, 05:09 PM
  4. Replies: 6
    Last Post: January 31st 2010, 12:17 PM
  5. derivatives involving square roots
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 31st 2009, 06:38 PM

Search Tags


/mathhelpforum @mathhelpforum