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I was so lost on this limits question. help please

  1. Oct 11, 2005 #1
    I just had a midterm for my calc class and there was this one question that I spent a good 20 minutes on and could not figure out. can someone help me with the solution please?
    lim x→0 (sin x²)/sin²x

    I can do it using L'Hopitals rule, but that wasn't allowed :uhh:
  2. jcsd
  3. Oct 11, 2005 #2


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    lim x→0 (sin x²)/sin²x = lim x→0 [(sin x²)/x²][x²/sin²x]
  4. Oct 11, 2005 #3


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    Use the Maclaurin's series for sinx and cosx.
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