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Math Help - help with a limit

  1. #1
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    help with a limit

    Hi
    I need a bit of help here....not sure how to start with
    lim x->0 (cos(^2)x - cosx)/x

    if i take cosx out [cosx(cosx - 1)]/x
    doesn't that give me cosx/x out the front which is undefined?....or am I getting all turned around?

    Sorry about how messy this is....I'm still learning how to use latex

    Thanks every one
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  2. #2
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    Quote Originally Posted by SimonK View Post
    Hi
    I need a bit of help here....not sure how to start with
    lim x->0 (cos(^2)x - cosx)/x

    if i take cosx out [cosx(cosx - 1)]/x
    doesn't that give me cosx/x out the front which is undefined?....or am I getting all turned around?

    Sorry about how messy this is....I'm still learning how to use latex

    Thanks every one
    Since this tends to \frac{0}{0} you can use L'Hospital's Rule.

    \lim_{x \to 0}\frac{\cos^2{x} - \cos{x}}{x}

     = \lim_{x \to 0}\frac{\frac{d}{dx}(\cos^2{x} - \cos{x})}{\frac{d}{dx}(x)}

     = \lim_{x \to 0}\frac{-2\sin{x}\cos{x} + \sin{x}}{1}

     = \lim_{x \to 0}\,(\sin{x} - 2\sin{x}\cos{x})

     = 0.
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  3. #3
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    Thanks very much
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  4. #4
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    another way

    Quote Originally Posted by SimonK View Post
    Thanks very much
    cosx(cosx-1)/x
    =cosx(2sin^2(x/2))x/x^2
    =sin part becomes 1 and cos part becomes 0
    =0 hence the answer.
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