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Math Help - How do I solve these?

  1. #1
    Junior Member ImaCowOK's Avatar
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    How do I solve these?

    I tried them several times and got them wrong. I don't understand the way my prof explains how to do them either.

    If you could show step by step, as simple as possible, so that I can understand and be able to do them myself.

    \lim_{x \to 0}\frac{1-cosx}{x} lim does not exist

    and

    \lim_{x \to 0}\frac{tanx}{sin8x} lim is \frac{1}{8}

    \lim_{x \to 0} 7x\cdot cotx\cdot cscx lim is \frac{7}{11}

    I already have the correct answer - I just don't understand how to get there.
    Last edited by ImaCowOK; September 5th 2010 at 08:00 AM.
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  2. #2
    MHF Contributor
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    Quote Originally Posted by ImaCowOK View Post
    I tried them several times and got them wrong. I don't understand the way my prof explains how to do them either.

    If you could show step by step, as simple as possible, so that I can understand and be able to do them myself.

    Limit of \frac{1-cosx}{x} as x -> 0 lim does not exist

    and

    Lim of \frac{tanx}{sin8x} as x -> 0 lim is \frac{1}{8}

    I already have the correct answer - I just don't understand how to get there.
    \displaystyle \frac{1-\cos{x}}{x^3} \cdot \frac{1+\cos{x}}{1+\cos{x}} = \frac{1-\cos^2{x}}{x^3(1+\cos{x})} = \frac{\sin^2{x}}{x^3(1+\cos{x})} = \frac{\sin^2{x}}{x^2} \cdot \frac{1}{1+\cos{x}} \cdot \frac{1}{x}<br />

    now evaluate ...

    \displaystyle \lim_{x \to 0} \frac{\sin^2{x}}{x^2} \cdot \frac{1}{1+\cos{x}} \cdot \frac{1}{x}


    for the second limit ...

    \displaystyle \frac{\tan{x}}{\sin(8x)} \cdot \frac{8x}{8x} = \frac{8x \tan{x}}{8x \sin(8x)} = \frac{1}{8} \cdot \frac{\tan{x}}{x} \cdot \frac{8x}{\sin(8x)}

    now evaluate ...

    \displaystyle \lim_{x \to 0} \frac{1}{8} \cdot \frac{\tan{x}}{x} \cdot \frac{8x}{\sin(8x)}
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  3. #3
    Junior Member ImaCowOK's Avatar
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    Quote Originally Posted by skeeter View Post
    \frac{\tan{x}}{x}
    How did you get that? I got \frac{\tan{x}}{8x} instead on that step after = \frac{8x \tan{x}}{8x \sin(8x)}
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  4. #4
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    Quote Originally Posted by ImaCowOK View Post
    How did you get that? I got \frac{\tan{x}}{8x} instead on that step after = \frac{8x \tan{x}}{8x \sin(8x)}
    \displaystyle \frac{\tan{x}}{8x} = \frac{1}{8} \cdot \frac{tan{x}}{x}
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  5. #5
    Junior Member ImaCowOK's Avatar
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    Okay, I get it.
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