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Can't understand how to compute this limit where x tends to infinity

  1. Jan 3, 2013 #1
    1. The problem statement, all variables and given/known data

    This is a question from a past exam paper:

    2. Relevant equations



    3. The attempt at a solution

    I really had no idea how to approach this but the solution is:
    Hopefully someone can explain to me the method used to obtain this answer.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 3, 2013 #2

    HallsofIvy

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    It's a fairly standard technique- "infinity" is hard to deal with while "0" is very easy- so they have made use of the fact if x goes to infinity, 1/x goes to 0.

    Starting from [itex](2x^3+ x)/(3x^2- 4x^3)[/itex], they have divided both numerator and denominator by the highest power of x, [itex]x^3[/itex], to get [itex](2+ 1/x^2)/(3/x- 4)[/itex] and then replaced "1/x" with "y": (2+y^2)/(3y- 4). Then, as x goes to infinity, y goes to 0 so that the limit is just (2+ 0)/(0- 4)= 2/(-4)= -1/2.
     
    Last edited by a moderator: Jan 3, 2013
  4. Jan 3, 2013 #3
    Or at infinity you take the highest power in the numerator over the highest power in the denominator. In your case it would be 2x^3/-4x^3 and then x^3 will cancel out leaving you with -2/4 which is -1/2.
     
  5. Jan 3, 2013 #4
    Thanks very much for that - very clear.

    Thank you
     
    Last edited by a moderator: Jan 3, 2013
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