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Cosx math problem

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

    lim x->0 (1-cosx) all over x

    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


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    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




    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


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    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


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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


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

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