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Finding limit help.

  1. Oct 17, 2004 #1

    I was wondering if anyone here can help me with the following question.

    lim 1 - cos(x³ sin x) / x^6 sin² x

    I have tried multiplying the top and the bottom by 1 + cos(x³ sin x), so I can get

    (sin²(x³ sin x) / x^6 sin² x) * 1 / 1 + cos(x³ sin x)

    I get stuck there. I don't know what to do next. Any help would be great thanks.
  2. jcsd
  3. Oct 17, 2004 #2


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    Welcom to PF!
    Just to make sure I understand your notation, you are seeking:

    Use the sneaky [tex]\sin(y)\approx{y},\cos(y)\approx1 (y\approx0)[/tex] to find the limit.
    You should end up with [tex]\frac{1}{2}[/tex]
  4. Oct 17, 2004 #3
    sin^2(x^3sin(x)) / (x^3sin(x)) * 1 / (x^3sin(x)) * 1 / 1 + cos(x^3sin(x))

    The limit of the first one is 1 (since the argument of the sin and denominator are the same) and the limit of the last is 1/2 (cos0 is 1 so that is 1/2)

    that is all about I know ^^

    The middle part poses a problem for my limited experience ><
  5. Oct 17, 2004 #4
    Thanks for the reply. I am currently taking first-year Calculus in University. It is very difficult.

    I do not understand what the [tex]\sin(y)\approx{y},\cos(y)\approx1 (y\approx0)[/tex] notation is. I have never seen it before, unless I am missing something.
  6. Oct 17, 2004 #5
    Nice eye singleton. I didn't even notice the part of how I could have broken up x^6 sin x

    Thanks for the help. I'll try it out.
  7. Oct 17, 2004 #6


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    What it means, is that when the argument of sine (y) is approximately zero, then
    sin(y) is approximately equal to y.
    You've seen the following limit before:
    [tex]\lim_{x\to0}\frac{\sin(x)}{x}=1[/tex] (Right?)
    This implies that [tex]\sin(x)\approx{x}[/tex] when x is close to zero..
  8. Oct 17, 2004 #7
    nevermind I'm wrong! (can't seem to get rid of an undefined anyway I try and cut it) try to google the method the other poster suggested!
    Last edited: Oct 17, 2004
  9. Oct 17, 2004 #8
    Thanks for the help you two!
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