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Math Help - [SOLVED] another trig limit question

  1. #1
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    [SOLVED] another trig limit question

    This one is throwing me for a loop:
    \lim_{x\to 0}\frac{1-cosx}{x}=0
    how do I show this using the fact that \lim_{x\to 0}\frac{sinx}{x}=1?
    I realize this is more a question of manipulating the function using trig identities, but I really cant see it.
    Any help?
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  2. #2
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    Quote Originally Posted by dannyboycurtis View Post
    This one is throwing me for a loop:
    \lim_{x\to 0}\frac{1-cosx}{x}=0
    how do I show this using the fact that \lim_{x\to 0}\frac{sinx}{x}=1?
    I realize this is more a question of manipulating the function using trig identities, but I really cant see it.
    Any help?
    \lim_{x\to 0}\frac{1-\cos x}{x}= \lim_{x\to 0}\frac{(1- \cos x)(1 + \cos x)}{x(1 + \cos x)}

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

    =  \lim_{x\to 0}\frac{\sin x}{x} \cdot \lim_{x\to 0}\frac{\sin x}{(1 + \cos x)}.
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