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Limit of a function - Indeterminate Form and LH rule

  1. Feb 28, 2010 #1
    1. The problem statement, all variables and given/known data


    Lim x-> 0 Sin(x) + Cos(x) + e^-x - 2 / Sin(x) - x



    3. The attempt at a solution

    Sin(x) + Cos(x) + e^-x - 2 = 0 and Sin(x) - x = 0

    This is indeterminate form?

    I used L'Hopitals rule 3 times to get

    -Cos(x) + Sin(x) - e^-x / -Cos(x)

    The top equals -2 and the bottom equals -1. The limit is 2. Am I understanding this right?

    1. The problem statement, all variables and given/known data

    Lim x->infinity (sqrt(x^2+1) - 1)^ln(x)

    3. The attempt at a solution

    I'm not sure what to do with ln(x). I believe this is the indeterminate form infinity - infinity.
    Am I supposed to rewrite this someway?
     
  2. jcsd
  3. Feb 28, 2010 #2
    should be -sinx, not +sinx after doing it three times.
    But yes, the answer is 2.
     
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