Limit of trigonometric function

[tmt1]

I have this problem:

$\lim_{{t}\to{0}} \frac{tan(6t)}{sin2t}$

I know sin2t = 0 when t = 0, which means the original fraction is indeterminate, so how can apply the rules for limits to solve this limit?

 

[ineedhelpnow]

hello there! do you know how to use L'Hospitals rule? i recommend you start with that. you can then easily simplify what you get from that (becomes extremely straight forward and simple after this) and then just take the limit of the top and bottom as x goes to 0.

 

[soroban]

Hello, tmt!

We know that: .$\displaystyle\lim_{x\to0}\frac{\sin x}{x} \,=\,1$

$\displaystyle \lim_{t\to0} \frac{\tan(6t)}{\sin(2t)}$

$\displaystyle \frac{\tan(6t)}{\sin(2t)} \;=\;\frac{\frac{\sin(6t)}{\cos(6t)}}{\sin(2t)} \;=\;\frac{\sin(6t)}{\sin(2t)\cos(6t)}\;=\; \frac{\frac{6t}{6t}\cdot\sin(6t)}{\frac{2t}{2t}\cdot\sin(2t)\cos(6t)} $

$\displaystyle\qquad=\;\frac{6t\cdot\frac{\sin(6t)}{6t}}{2t\cdot\frac{\sin(2t)}{2t}}\;=\; 3\cdot\frac{\frac{\sin(6t)}{6t}}{\frac{\sin(2t)}{2t}\cdot\cos(6t)} $$\displaystyle \lim_{t\to0}\left( 3\cdot\frac{\frac{\sin(6t)}{6t}}{\frac{\sin(2t)}{2t}\cdot\cos(6t)}\right) \;=\; 3\cdot\frac{\lim \frac{\sin(6t)}{6t}}{\lim\frac{\sin(2t)}{2t}\cdot \lim\cos(6t)} \;=\;3\cdot\frac{1}{1\cdot1} \;=\;3$