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Math Help - calculating limits invloving trig functions

  1. #1
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    calculating limits invloving trig functions

    Here is the problem.

    lim as x approaches 0 of (tan4x/sin2x)

    So far I have done this.

    (sin7x/cos7x)/(sin2x)

    (sin7x/cos7x) multiplied by (1/sin2x)

    I'm not sure where to go from here. Suggestions? Thanks.

    juventinoalex
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  2. #2
    Junior Member bondesan's Avatar
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    Try to use the fundamental trigonometric limit \lim_{x\to 0}\dfrac{sin(x)}{x} = 1 and its consequences when it comes to calculate the limit for the tangent.
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  3. #3
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    I still don't think I understand. Could someone explain further?

    Thanks,
    juventinoalex
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  4. #4
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    1) Is it tan(4x) or tan(7x)?

    2) \frac{sin(7x)}{sin(2x)cos(7x)} = sin(7x) \cdot \frac{7x}{7x} \cdot \frac{1}{sin(2x)} \cdot \frac{2x}{2x} \cdot \frac{1}{cos(7x)}  = \frac{7}{2} \cdot \frac{sin(7x)}{7x} \cdot \frac {2x}{sin(2x)} \cdot \frac{1}{cos(7x)}
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  5. #5
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    Quote Originally Posted by juventinoalex View Post
    I still don't think I understand. Could someone explain further?
    \displaystyle \tan(4x)=\frac{\sin(4x)}{\cos(4x)}=\frac{2\sin(2x)  \cos(2x)}{\cos(4x)}.

    So \displaystyle \frac{\tan(4x)}{\sin(2x)}=\frac{2\cos(2x)}{\cos(4x  )}}
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  6. #6
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    Sorry for the typo. It is tan4x
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  7. #7
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    See my reply #5.
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  8. #8
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    Got it now. Thanks so much.
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