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: Epsilon-delta proof of limits

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove (using epsilon-delta definition only) that the limit of the following expression:
    is 1/2, as n tends towards infinity.

    2. Relevant equations
    For any ε > 0, there exists some natural number N, such that:
    n > N gives|f(n) - L| < ε

    3. The attempt at a solution
    Multiply by the conjugate pair and simplify to obtain:

    Taking the expression to be compared against ε
    |[itex]n/[sqrt(n^2+n)+n] - 1/2[/itex]| and finding a common denominator gives:

    [itex]|2n/2[sqrt(n^2+n)+n] - [sqrt(n^2+n)+n]/2[sqrt(n^2+n)+n]|[/itex]
    [itex]|[n-sqrt(n^2+n)] / 2[sqrt(n^2+n)+n]|[/itex]

    The numerator is always negative and the denominator is always positive, so the expression can be re-written as:
    [itex][sqrt(n^2+n) - n][/itex] / [itex]2[sqrt(n^2+n)+n][/itex]

    I chose to split this up into two terms:
    [itex]sqrt(n^2+n)[/itex] / [itex]2[sqrt(n^2+n)+n][/itex] minus
    [itex]n[/itex] / [itex]2[sqrt(n^2+n)+n][/itex]

    I have an expression in the form a/b-c/d. Decreasing "b", decreasing "c" and increasing "d" both increase the value of the expression, thus the expression below is strictly greater than the
    expression above.

    [itex]sqrt(n^2+n)[/itex] / [itex]2[sqrt(n^2+n)][/itex] minus
    [itex]1[/itex] / [itex]2[sqrt(n^2+n^2)+n][/itex]

    This simplifies to [itex]1/2[/itex] - [itex]1[/itex] / [itex]n(2sqrt(2)+1)][/itex]

    Both terms in this expression are positive. Thus, changing the subtraction to addition will strictly increase the value of the entire quantity.

    Thus, comparing
    [itex]1/2[/itex] + [itex]1[/itex] / [itex]n(2sqrt(2)+1)[/itex] against ε is enough.

    Solving for n results in some mess that results in:
    n > (positive constant) / (ε - 1/2)

    Clearly, something has gone awry. Consideration of an arbitrarily small ε reveals the RHS to be negative. That would imply that the first term in my sequence is already arbitrarily close to my limit.

    Where did I go wrong?
  2. jcsd
  3. Feb 7, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    I'm not sure. Why don't you just take your result n/(sqrt(n^2+1)+n) and write it as n/(n*sqrt(1+1/n^2)+n), cancel the n's and give it a fresh thought.
  4. Feb 7, 2009 #3


    Staff: Mentor

    Here's an approach that's simpler, that I think you can use.
    You can start with this inequality:
    [tex]\sqrt{n^2 + n} - n < \epsilon[/tex]

    You don't need absolute values, since the radical expression is always > n for n > 0, hence the expression on the left side above is always positive.

    Now, add n to both sides, and then square both sides, the idea being that if a < b, then a^2 < b^2, where a and b are positive.

    That gets you to
    [tex]n^2 + n < (n + \epsilon)^2[/tex]

    After cleaning a bit, you get to
    [tex]\epsilon^2 + 2n\epsilon - n > 0[/tex]

    That's a quadratic in epsilon, so that should get you a relationship between epsilon and n, from which you can find the number N that you want.
  5. Feb 8, 2009 #4


    User Avatar
    Science Advisor
    Homework Helper

    Uh, you don't want to show |sqrt(n^2+n)-n|<e. It isn't. The limit is 1/2. You want to show |sqrt(n^+n)-n-1/2|<e.
  6. Feb 8, 2009 #5


    Staff: Mentor

    Oops! Well, no wonder it was easier. After my first stab got nowhere, I used a clean sheet of paper and omitted that 1/2.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook