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Limit as x approaches 0.

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data
    http://www.prep101.com/files/Math100PracticeExam.pdf [Broken]

    Question 1m

    3. The attempt at a solution
    I tried to do this by using
    lim sinx/x = 1
    x->0

    Factoring out x^2 from each one and I get infinity.
    Why does this method not work?

    The answer says it's 4/5.

    I even tried plugging in 0.001 in my calculator and I get infinity :S.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 8, 2011 #2
    Is that sin(2x^2), sin^2(2x), sin(2)x^2 ?
     
  4. Nov 8, 2011 #3

    Dick

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    Or is it 3sin(2x^2-6x^2+4x^6)? The question isn't very grammatical. And however I rearrange it I still can't figure out how to pull 4/5 as a limit out.
     
  5. Nov 8, 2011 #4
    It is 3sin(2x^2), you get 4/5, this limit is INSANE :P
     
  6. Nov 8, 2011 #5

    vela

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    It helps to use the substitution u=x2 to get
    [tex]\lim_{u \to 0}\frac{3 \sin 2u - 6u + 4u^3}{u^5}[/tex]Then three applications of the Hospital rule gets you to a limit you can evaluate.
     
  7. Nov 9, 2011 #6

    Dick

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    Ah, indeed it does.
     
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