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Math Help - Trig Limit

  1. #1
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    Trig Limit

    I'm really struggling with these trig limits. I have three of them that I can't get straight, which I'll post separately. Sorry in advance for dropping all of these. I've stared at them endlessly, and don't know what else to do. Here's the first one:

    \displaystyle \lim_{x\to 0 } \frac {sin 4x}{sin 6x}

    I want to say that \frac {sin 4x}{sin 6x} = \frac{2}{3} sin x, but I know this isn't correct. The answer to this is \frac{2}{3}, but I have no idea how to get there. We haven't covered L'Hospitals rule yet, so somehow I think that I have to get this down to \frac {2}{3} * \frac {sin x}{x}, but I just can't figure it out.

    Can somebody please help?
    Thanks.
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  2. #2
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    Quote Originally Posted by reastland View Post
    I'm really struggling with these trig limits. I have three of them that I can't get straight, which I'll post separately. Sorry in advance for dropping all of these. I've stared at them endlessly, and don't know what else to do. Here's the first one:

    \displaystyle \lim_{x\to 0 } \frac {sin 4x}{sin 6x}

    I want to say that \frac {sin 4x}{sin 6x} = \frac{2}{3} sin x, but I know this isn't correct. The answer to this is \frac{2}{3}, but I have no idea how to get there. We haven't covered L'Hospitals rule yet, so somehow I think that I have to get this down to \frac {2}{3} * \frac {sin x}{x}, but I just can't figure it out.

    Can somebody please help?
    Thanks.
    Can you use L'Hopitals Rule?

    Never mind. I just read you haven't covered it.

    Try double angle formulas.
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  3. #3
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    Quote Originally Posted by reastland View Post

    I want to say that \frac {sin 4x}{sin 6x} = \frac{2}{3} sin x, but I know this isn't correct.
    It's not correct.

    Have you tried to use double angle formulas?
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  4. #4
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    \displaystyle \frac{\sin{4x}}{\sin{6x}} = \frac{2\sin{2x}\cos{2x}}{2\sin{3x}\cos{3x}}

    \displaystyle = \frac{\sin{2x}\cos{2x}}{\sin{3x}\cos{3x}}

    \displaystyle = \frac{2\sin{x}\cos{x}\cos{2x}}{\sin{x}(3 - 4\sin^2{x})\cos{3x}}

    \displaystyle = \frac{2\cos{x}\cos{2x}}{(3 - 4\sin^2{x})\cos{3x}}.


    You should now be able to evaluate the limit.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by reastland View Post
    I'm really struggling with these trig limits. I have three of them that I can't get straight, which I'll post separately. Sorry in advance for dropping all of these. I've stared at them endlessly, and don't know what else to do. Here's the first one:

    \displaystyle \lim_{x\to 0 } \frac {sin 4x}{sin 6x}

    I want to say that \frac {sin 4x}{sin 6x} = \frac{2}{3} sin x, but I know this isn't correct. The answer to this is \frac{2}{3}, but I have no idea how to get there. We haven't covered L'Hospitals rule yet, so somehow I think that I have to get this down to \frac {2}{3} * \frac {sin x}{x}, but I just can't figure it out.

    Can somebody please help?
    Thanks.
    If you want to use \lim\limits_{x\to0}\dfrac{\sin x}{x}=1, we can think of your problem as:

    \lim\limits_{x\to0}\dfrac{\sin(4x)}{\sin(6x)}=\lim  \limits_{x\to0}\dfrac{4\dfrac{\sin(4x)}{4x}}{6\dfr  ac{\sin(6x)}{6x}}=\dfrac{4\lim\limits_{x\to0}\dfra  c{\sin(4x)}{4x}}{6\lim\limits_{x\to0}\dfrac{\sin(6  x)}{6x}}

    Now can you see why the solution is \dfrac{2}{3}?
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  6. #6
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    Chris, you are awesome. Thanks a million!
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