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Math Help - [SOLVED] Limit of Trigonometric Functions

  1. #1
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    [SOLVED] Limit of Trigonometric Functions

    Find the limit of tan(6t)/sin(2t) as x---> 0
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  2. #2
    o_O
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    We're going to play around with the expression a little bit so that we can use this: \lim_{u \to 0} \frac{\sin u}{u} = \lim_{u \to 0} \frac{u}{\sin u} = 1

    With your limit:
    \begin{aligned} \lim_{t \to 0} \frac{\tan (6t)}{\sin (2t)} & = \lim_{t \to 0} \frac{\sin (6t)}{\cos (6t) \cdot \sin (2t)} \\ & = \lim_{t \to 0} \left(\sin (6t) \cdot \frac{1}{\sin (2t)} \cdot \frac{1}{\cos (6t)} \right) \end{aligned}

    Now, somehow we're going to use the very first limit to get rid of the \sin terms. We can do so by ...

    \lim_{t \to 0} \left(\sin (6t) \cdot {\color{red}\frac{6t}{6t}} \cdot \frac{1}{\sin (2t)} \cdot {\color{blue} \frac{2t}{2t}} \cdot \frac{1}{\cos (6t)}\right)

     = \lim_{t \to 0} \left( \frac{\sin (6t)}{{\color{red}6t}} \cdot \frac{{\color{blue}2t}}{\sin (2t)} \cdot \frac{{\color{red}6t}}{{\color{blue}2t} \cos (6t)} \right)

    = \underbrace{\lim_{t \to 0} \frac{\sin (6t)}{{\color{red}6t}}}_{=1} \ \cdot \ \underbrace{ \lim_{t \to 0} \frac{{\color{blue}2t}}{\sin (2t)}}_{=1} \ \cdot \ \lim_{t \to 0}\frac{{\color{red}6t}}{{\color{blue}2t} \cos (6t)}

    = \lim_{t \to 0} \frac{3}{\cos (6t)}

    where you can directly sub in t=0
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  3. #3
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    oh boy..........

    Quote Originally Posted by o_O View Post
    We're going to play around with the expression a little bit so that we can use this: \lim_{u \to 0} \frac{\sin u}{u} = \lim_{u \to 0} \frac{u}{\sin u} = 1

    With your limit:
    \begin{aligned} \lim_{t \to 0} \frac{\tan (6t)}{\sin (2t)} & = \lim_{t \to 0} \frac{\sin (6t)}{\cos (6t) \cdot \sin (2t)} \\ & = \lim_{t \to 0} \left(\sin (6t) \cdot \frac{1}{\sin (2t)} \cdot \frac{1}{\cos (6t)} \right) \end{aligned}

    Now, somehow we're going to use the very first limit to get rid of the \sin terms. We can do so by ...

    \lim_{t \to 0} \left(\sin (6t) \cdot {\color{red}\frac{6t}{6t}} \cdot \frac{1}{\sin (2t)} \cdot {\color{blue} \frac{2t}{2t}} \cdot \frac{1}{\cos (6t)}\right)

     = \lim_{t \to 0} \left( \frac{\sin (6t)}{{\color{red}6t}} \cdot \frac{{\color{blue}2t}}{\sin (2t)} \cdot \frac{{\color{red}6t}}{{\color{blue}2t} \cos (6t)} \right)

    = \underbrace{\lim_{t \to 0} \frac{\sin (6t)}{{\color{red}6t}}}_{=1} \ \cdot \ \underbrace{ \lim_{t \to 0} \frac{{\color{blue}2t}}{\sin (2t)}}_{=1} \ \cdot \ \lim_{t \to 0}\frac{{\color{red}6t}}{{\color{blue}2t} \cos (6t)}

    = \lim_{t \to 0} \frac{3}{\cos (6t)}

    where you can directly sub in t=0
    That's a tough one. Thanks!
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