# Thread: [SOLVED] Limit of Trigonometric Functions

1. ## [SOLVED] Limit of Trigonometric Functions

Find the limit of tan(6t)/sin(2t) as x---> 0

2. We're going to play around with the expression a little bit so that we can use this: $\lim_{u \to 0} \frac{\sin u}{u} = \lim_{u \to 0} \frac{u}{\sin u} = 1$

\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 $t=0$

3. ## oh boy..........

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

\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!