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Math Help - the limit of tan6t/sin2t as t approaches zero

  1. #1
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    the limit of tan6t/sin2t as t approaches zero

    the limit of tan6t/sin2t as t approaches zero-screen-shot-2012-10-23-12.53.53-am.png


    I understand after the first step, but not how the multiplication was done to go from tan6t/sin2t to the next step. When I did it I left out the t in the denominator of sin6t/t and the t in the numerator of t/sin2t. I just had sin6t/1 and 1/sin2t
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    Lightbulb Re: the limit of tan6t/sin2t as t approaches zero

    Picture shows a dumb approach to finding the limit.

    Just expand Tan(6t) and sin(2t) and simplify the ratio to:

    \frac{\tan (6 t)}{\sin (2 t)} = (2 \cos (4 t)+1) \sec (6 t)

    \lim_{t\to 0} \, (2 \cos (4 t)+1) \sec (6 t) = 3
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    Re: the limit of tan6t/sin2t as t approaches zero

    Quote Originally Posted by kingsolomonsgrave View Post
    Click image for larger version. 

Name:	Screen Shot 2012-10-23 at 12.53.53 AM.png 
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    I understand after the first step, but not how the multiplication was done to go from tan6t/sin2t to the next step. When I did it I left out the t in the denominator of sin6t/t and the t in the numerator of t/sin2t. I just had sin6t/1 and 1/sin2t
    Yes, if you "leave out" the "t"s you have hust sin(6t)/1 and 1/sin(2t) but that doesn't help! If you then take the limit as t goes to 0 you have sin(6t)0\to 0 and 1/sin(2t)\to \infty which leads to 0\cdot\infty- and that doesn't give anything. What they are doing is using the basic limit, \lim_{x\to 0}\frac{sin(x)}{x}= 1 with x= 6t in the first case and x= 2t in the second.
    Thanks from topsquark
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