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Math Help - Trigonometry Limits

  1. #1
    Newbie
    Joined
    Apr 2010
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    Trigonometry Limits

    I have this problem:

    lim (x-->1) sin(2x)/3x

    I multiplied the whole thing by 2/2 and took a 2/3 out of the limit, so I then have:
    2/3 [lim(x-->1) sin(2x)/2x

    Which of course is very similar to the sinx/x limit, but the variable is approaching 1, not 0 in this case...

    Would it be correct to do this:
    Let z = x-1
    x-->1, then z-->0
    and turn the equation into:
    (2/3) [lim(z-->0) sin2(z+1)/2(z+1)
    I feel like this isn't going to get me anywhere useful...

    Any tips, please? I appreciate it.
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  2. #2
    Junior Member
    Joined
    Mar 2010
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    The function \frac{\sin(2x)}{3x} is continuous at 1 so \lim_{n \to 1} \frac{\sin(2x)}{3x} = \frac{\sin(2 \cdot 1)}{3 \cdot 1} = \frac{\sin(2)}{3}.
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  3. #3
    Newbie
    Joined
    Apr 2010
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    Quote Originally Posted by Giraffro View Post
    The function \frac{\sin(2x)}{3x} is continuous at 1 so \lim_{n \to 1} \frac{\sin(2x)}{3x} = \frac{\sin(2 \cdot 1)}{3 \cdot 1} = \frac{\sin(2)}{3}.
    So no fancy manipulation to make it into that trig limit format? Wow... nice and simple. Thanks Giraffo!
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