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Math Help - Trig Limits

  1. #1
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    Trig Limits

    Hi
    I'm having serious trouble with trig limits and I'm wondering if anyone could explain it to me in simple terms as I don't really understand the textbook I have. Any help would be greatly appreciated.

    For example in my textbook there is an example however I don't understand it. Could any help and just talk us through it?

    lim x approaches 0 (1 - cos x)/x = lim x app 0 ((1 - cos x) / x (1 + cos x)/ 1 + cos x)
    I don't understand where the 1 + cos x divided by x came from
    then it goes on
    = 1 - cos^2 x / x(1 + cos x)
    = sin^2 x/ x (1 + cos x)
    = (sin x/x * sinx x /(1 + cos x))
    = sinx x * lim x app 0 sin x/(1 + cos x)
    = 1 * sin 0/1 + cos 0
    = 0

    I can't follow it through so if anyone can explain the steps and tell me how you get that it would ease my troubles. Thanks.
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  2. #2
    Super Member wingless's Avatar
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    \lim_{x\to 0} \frac{1-\cos x}{x}

    \lim_{x\to 0} \frac{1-\cos x}{x}\cdot \frac{1+\cos 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)}

    As \lim_{x\to 0} \frac{\sin x}{x} = 1,

    \lim_{x\to 0} \frac{\sin^{\not 2} x}{\not{x}(1+\cos x)}

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

    Now put x=0,

    \frac{\sin 0}{1+\cos 0}

    \frac{0}{2} = 0


    That \frac{1+\cos x}{1+\cos x} didn't come from anywhere, we multiplied \frac{1-\cos x}{x} by it to get a solution. To see such things you have to be familiar with trigonometric limits and identities.
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  3. #3
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    Hello, Mickey!

    A slightly different explanation . . .

    We're expected to know that: . \lim_{x\to0}\frac{\sin x}{x} \:=\:1


    \lim_{x\to0} \frac{1-\cos x}{x}

    Multiply by \frac{1+\cos x}{1 + \cos x}\!:\;\;\;\frac{1-\cos x}{x}\cdot{\color{blue}\frac{1+\cos x}{1 + \cos x}} \;\;=\;\;\frac{1-\cos^2x}{x(1+\cos x)} \;\;=\;\;\frac{\sin^2x}{x(1+\cos x)}


    And we have: . \lim_{x\to0}\left[\frac{\sin x}{x}\cdot\frac{\sin x}{1 + \cos x}\right] \;\;=\;\;1\cdot\frac{0}{1+1} \;\;=\;\;0

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