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Math Help - Limit of a trig functions

  1. #1
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    Limit of a trig functions

    hi, can someone show me why sin(3x)/2x = 3/2? is it because they multiplied sinx/x by 3/2? if so why does 3sinx = sin3x??
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  2. #2
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    Quote Originally Posted by Arez View Post
    hi, can someone show me why sin(3x)/2x = 3/2? is it because they multiplied sinx/x by 3/2? if so why does 3sinx = sin3x??
    \lim_{x \rightarrow 0}\frac{\sin (3x)}{2x} = \lim_{x \rightarrow 0} \left( \frac{3}{2} \cdot \frac{\sin (3x)}{3x} \right) = \frac{3}{2} \cdot \lim_{t \rightarrow 0}\frac{\sin t}{t} where t = 3x ....
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  3. #3
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    on that note then, how would I do sin^2(3t)/2t? the squared is throwing me off..
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    First off I think a new thread is in order but...

     \lim_{x \to 0} \frac{\sin^2 (3x)}{2x} = \left( \lim_{x \to 0} \sin(3x) \right) \cdot \left( \lim_{x \to 0} \frac{\sin(3x)}{2x} \right) \; \; ...
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