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Limit from some russian book: lmt-0 [(1+mx)^n-(1+nx)^m]/x^2

  1. Jul 13, 2010 #1
    limit from some russian book: lmt--0 [(1+mx)^n-(1+nx)^m]/x^2

    Hey guys, i'm newbie in calculus, while doing a quick reading of my problem's book i find this interesting limit:
    1. The problem statement, all variables and given/known data
    solve the following limit when x tend to zero,
    m and n belong to the set of naturals
    [tex]\frac{(1+mx)^n-(1+nx)^m}{ x^2}[/tex]


    2. Relevant equations

    3. The attempt at a solution
    the anwser is (mn)(m-n)
    i dont know how to solve the problem...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 13, 2010 #2

    Mark44

    Staff: Mentor

    Re: limit from some russian book: lmt--0 [(1+mx)^n-(1+nx)^m]/x^2

    The approach I would take would be to expand both binomials using the binomial theorem http://en.wikipedia.org/wiki/Binomial_theorem.
     
  4. Jul 13, 2010 #3
    Re: limit from some russian book: lmt--0 [(1+mx)^n-(1+nx)^m]/x^2

    or use L'Hospital's rule (twice).
    also.. your answer isn't right.
     
  5. Jul 13, 2010 #4

    Mark44

    Staff: Mentor

    Re: limit from some russian book: lmt--0 [(1+mx)^n-(1+nx)^m]/x^2

    I like your answer better than mine.
     
  6. Jul 13, 2010 #5
    Re: limit from some russian book: lmt--0 [(1+mx)^n-(1+nx)^m]/x^2

    Exactly!
    Before posting, i've tryed L'hopp, and i reached the following answer:
    (mn)(m-n)(1/2)
    Right?, but looking at the answer's section in the book, i realized that the answer given by the book is:
    (mn)(m-n)

    Warning! my silly idea:
    expanding the binomial, and doing the substitution of all x, almost every term of the function will be zero, it doesn't matter the value of m and n, am i wrong? What do you think about it?
    i appreciate all your comments and if you notice any mistakes in my writing (Sintax, grammar, etc.), let me know*(?), English is not my mother tongue, but I want to learn it.
     
  7. Jul 13, 2010 #6
    Re: limit from some russian book: lmt--0 [(1+mx)^n-(1+nx)^m]/x^2

    When you expand with the binomial theorem, in the numerator, what is the coefficient of x? What happens when you divide that by x2 as x approaches zero?

    EDIT: Also, your idea isn't silly. Most ideas aren't silly but some lead to silly conclusions when you realize that you didn't get anywhere. And some leave you feeling silly since you didn't follow them to completion. But the idea, well, that's just being a creative person.
     
    Last edited: Jul 13, 2010
  8. Jul 14, 2010 #7
    Re: limit from some russian book: lmt--0 [(1+mx)^n-(1+nx)^m]/x^2

    Very nice comment Tedjn!! :cool:
     
  9. Jul 15, 2010 #8
    Re: limit from some russian book: lmt--0 [(1+mx)^n-(1+nx)^m]/x^2

    Alex Spivak ? :)
     
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