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When does the limit exist?

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Find a value of k such that the limit exists

    lim (x2-kx+4)/(x-1)
    x->1

    3. The attempt at a solution

    In the solution they set the top equal to zero finding that k is equal to 5. Why do you assume the numerator must equal zero if the denominator equals zero? Is this the only case in which the limit exists? If so why?
     
  2. jcsd
  3. Dec 9, 2009 #2

    ideasrule

    User Avatar
    Homework Helper

    If the top equals anything else, the fraction (x^2-kx+4)/(x-1) would blow up near x=1 because the denominator approaches 0. Setting the top to 0 ensures that x-1 is a factor of the numerator. The x-1 on the top would then cancel with the denominator, preventing the fraction from becoming infinity near x=1.
     
  4. Dec 9, 2009 #3
    So in this case, they are assuming that an infinite limit does not exist?
     
  5. Dec 10, 2009 #4

    Mark44

    Staff: Mentor

    They are implicitly saying that a finite limit exists.
     
  6. Dec 10, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I would not use the word "assuming" here- it is a fact.

    Saying that a limit "is infinity" or "is negative infinity" is just saying that the limit does not exist for a specific reason.
     
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