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Math Help - Trig Limits.

  1. #1
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    Trig Limits.

    Hi, I'm having a lot of trouble in solving the following limit problems (all of the, as x --> 0):

    1) \frac{sin 7x}{sin 4x}

    2) \frac{sin(cos x)}{sec x}



    I'm very confused with these two problems, and would greatly appreciate any help (I'm not allowed to differentiate the functions).


    Thanks
    Last edited by mr fantastic; June 5th 2010 at 04:40 PM. Reason: Edited post title.
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  2. #2
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    Quote Originally Posted by spoc21 View Post
    Hi, I'm having a lot of trouble in solving the following limit problems (all of the, as x --> 0):

    1) \frac{sin 7x}{sin 4x}

    2) \frac{sin(cos x)}{sec x}



    I'm very confused with these two problems, and would greatly appreciate any help (I'm not allowed to differentiate the functions).


    Thanks
    \lim_{x \to 0}\frac{\sin{7x}}{\sin{4x}} = \lim_{x \to 0}\frac{\sin{(3x + 4x)}}{\sin{4x}}

     = \lim_{x \to 0}\frac{\sin{3x}\cos{4x} + \cos{3x}\sin{4x}}{\sin{4x}}

     = \lim_{x \to 0}\left(\frac{\sin{3x}\cos{4x}}{\sin{4x}} + \cos{3x}\right)

     = \lim_{x \to 0}\left(\frac{\sin{3x}\cos{4x}}{\sin{4x}}\right) + \lim_{x \to 0}\,(\cos{3x})

     = 1 + \lim_{x \to 0}\left(\frac{\sin{3x}\cos{4x}}{\sin{4x}}\right)

     = 1 + \lim_{x \to 0}\left[\frac{(3\sin{x} - 4\sin^3{x})(1 - 2\sin^2{2x})}{2\sin{2x}\cos{2x}}\right]

     = 1 + \lim_{x \to 0}\left[\frac{\sin{x}(3 - 4\sin^2{x})(1 - 2\sin^2{2x})}{4\sin{x}\cos{x}\cos{2x}}\right]

     = 1 + \lim_{x \to 0}\left[\frac{(3 - 4\sin^2{x})(1 - 2\sin^2{2x})}{4\cos{x}\cos{2x}}\right]

     = 1 + \frac{3}{4}

     = \frac{7}{4}.
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  3. #3
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    Quote Originally Posted by spoc21 View Post
    Hi, I'm having a lot of trouble in solving the following limit problems (all of the, as x --> 0):

    1) \frac{sin 7x}{sin 4x}

    2) \frac{sin(cos x)}{sec x}



    I'm very confused with these two problems, and would greatly appreciate any help (I'm not allowed to differentiate the functions).


    Thanks
    \lim_{x \to 0}\frac{\sin{(\cos{x})}}{\sec{x}} = \lim_{x \to 0}\,[\cos{x}\sin{(\cos{x})}]

     = \cos{(0)}\sin{[\cos{(0)}]}

     = 1\cdot \sin{(1)}

     = \sin{(1)}.
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