# Thread: the limit of tan6t/sin2t as t approaches zero

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

I understand after the first step, but not how the multiplication was done to go from $\displaystyle tan6t/sin2t$ to the next step. When I did it I left out the t in the denominator of $\displaystyle sin6t/t$ and the t in the numerator of $\displaystyle t/sin2t$. I just had $\displaystyle sin6t/1$ and $\displaystyle 1/sin2t$

2. ## 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:

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

$\displaystyle \lim_{t\to 0} \, (2 \cos (4 t)+1) \sec (6 t) = 3$

3. ## Re: the limit of tan6t/sin2t as t approaches zero

Originally Posted by kingsolomonsgrave

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

,

,

,

,

### tan6t/sin2t

Click on a term to search for related topics.