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Homework Help: Cosx math problem

  1. Mar 30, 2005 #1
    Am i doing these right?

    lim x->0 (1-cosx) all over x
    =(1-cosx/x)(1+cosx/1+cosx)
    =(1-cos^2x)/x(1+cosx)
    =(sin^2x)/x(1+cosx)
    =x(sin^2x)/xx(1+cosx)
    =0/2
    =0

    are my steps right?
     
  2. jcsd
  3. Mar 30, 2005 #2
    Nevermind, thats indeterminate. Your denominator is still 0. You may have to use l'hopitals rule.
     
  4. Mar 30, 2005 #3

    AKG

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    =x(sin^2x)/xx(1+cosx)
    =0/2

    That step is not right. The top will be 0*(sin²0) which is 0, the bottom will be 0 * 0 * 2 which is 0, not 2.

    What you can do is recognize that:

    x(sin²x)/x²(1 + cosx)
    = [x/(1 + cosx)][sin²x/x²]
    = [x/(1 + cosx)](sinx/x)²

    The limit as x -> 0 of (sinx/x) exists, and you should know it, and the limit as x -> 0 of [x/(1 + cosx)] exists, and you should be able to easily compute that, so the limit of the product is the product of the limits. That way, you get (0/2)(1)². Perhaps this is how you got rid of the x² on the bottom, and hence went from

    =x(sin^2x)/xx(1+cosx)

    to

    =0/2

    but it's not clear that you did that. At least show your work. Also, make sure that when you write it, don't just drop the "lim x->0" part until you actually compute it.
     
  5. Mar 31, 2005 #4

    HallsofIvy

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    [tex]\frac{1-cos(x)}{x}\frac{1+cos(x)}{1+cos(x)}[/tex]
    [tex]=\frac{sin^2(x)}{x(1+cos(x))}[/tex]
    [tex]=\frac{sin(x)}{x}\frac{sin(x)}{1+cos(x)}[/tex]

    What exactly do you already know and can use?
    If you are allowed to use the well known limit: [tex]\frac{sin x}{x}[/tex] goes to 1 and the facts that sine and cosine are continuous, so that the second fraction goes to 0/2= 0, it follows that the limit is 0.
     
  6. Mar 31, 2005 #5

    Galileo

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    I see no reason why the steps are inorect. Gillgill probably multiplied top and bottom by x, in order to write:

    [tex]\frac{\sin^2 x}{x(1+cosx)}=\frac{x\sin^2 x}{x^2(1+cosx)}=\left(\frac{\sin x}{x}\right)^2\frac{x}{1+\cos x}[/tex]

    so you can use that: [itex]\frac{\sin x}{x}[/itex] goes to one and all three products of limits exist.

    [tex]\lim _{x \to 0}\left(\frac{\sin x}{x}\right)^2\frac{x}{1+\cos x}=(1)(1)(0)=0[/tex]

    EDIT: It doesn't show in the calculations though, nevermind.
     
    Last edited: Mar 31, 2005
  7. Mar 31, 2005 #6

    dextercioby

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    An elegant way would have been to use the first 2 terms from cosine's Taylor series around 0...:wink:

    Daniel.
     
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