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Limit of [x^2(3 + sinx)] / [(x + sinx)^2]

  1. Sep 17, 2012 #1
    1. The problem statement, all variables and given/known data
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    2. Relevant equations



    3. The attempt at a solution

    I expanded everything out and got [3x^2 + x^2*sin(x)] / [x^2 + 2sin(x) + sin^2(x)] and I tried splitting the problem but it didn't work because the denominator is always 0 when I try plugging in '0' for 'x.' I can't find any way to cancel out the denominator
     
  2. jcsd
  3. Sep 17, 2012 #2

    Dick

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    You can't really 'cancel out' the denominator. You could try l'Hopital's rule or substituting a series expansion of sin(x). Or factoring x^2 out of the denominator and using a known limit like sin(x)/x.
     
  4. Sep 17, 2012 #3
    Oops, sorry, I forgot to mention this is for an introductory Calc 1 course, so none of that l'Hopital stuff...

    But what do you mean about substituting a series expansion?
     
  5. Sep 17, 2012 #4

    Dick

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    I think my last suggestion about factoring x^2 out of the denominator is the best. I added that to my post in an edit. Try that.
     
  6. Sep 17, 2012 #5
    So then I get:

    [x^2(3 + sinx)] / [x^2(1 + 2sinx/x + sin^2x/x] and cancel out the x^2's:

    (3 + sinx) / (1 + 2sinx/x + sin^2x/x)

    Editl: Oh, I had made an algebra error. I got it now, thanks
     
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