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Math Help - help with special trig limits

  1. #1
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    help with special trig limits

    Hi i'm having a problem with this limit. It seems like it woul be simple to do...

    using special trig limits find...

    lim 1-cosx / x^2
    x->0

    i tried breaking it up into 1-cosx/x and 1/x but the 1/x is still undefined.

    thanks in advance.
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  2. #2
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    \frac{1-\cos{x}}{x^2} \cdot \frac{1+\cos{x}}{1+\cos{x}} =<br />

    \frac{1-\cos^2{x}}{x^2(1 + \cos{x})} = <br />

    \frac{\sin^2{x}}{x^2(1 + \cos{x})} =

    \frac{\sin{x}}{x} \cdot \frac{\sin{x}}{x} \cdot \frac{1}{1+\cos{x}}

    now take the limit.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    \frac{1-\cos{x}}{x^2} \cdot \frac{1+\cos{x}}{1+\cos{x}} =<br />

    \frac{1-\cos^2{x}}{x^2(1 + \cos{x})} = <br />

    \frac{\sin^2{x}}{x^2(1 + \cos{x})} =

    \frac{\sin{x}}{x} \cdot \frac{\sin{x}}{x} \cdot \frac{1}{1+\cos{x}}

    now take the limit.

    Thanks big time! It was that stupid trig identity!! I can never remember those things.
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  4. #4
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    Quote Originally Posted by mattboersema View Post
    Hi i'm having a problem with this limit. It seems like it woul be simple to do...

    using special trig limits find...

    lim 1-cosx / x^2
    x->0

    i tried breaking it up into 1-cosx/x and 1/x but the 1/x is still undefined.

    thanks in advance.

    another method is given in the following link which breaks up the
    (1-cosx) into half angle
    limit

    another method could be L'Hospitals rule.
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  5. #5
    Junior Member toraj58's Avatar
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    Lightbulb

    according to SANDWICH Theory we can replace

    1-\cos u

    with

    \frac{1}{2} u^2

    when we get ambigious mode of \frac{0}{0}

    so the menthiond limit can be changed to:

    \lim_ {x\to 0} \frac {\frac {1}{2} x^2}{x^2} = \frac {1}{2}
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