Question

use the relation $lim_{\theta
ightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit.
$lim_{t
ightarrow0}\frac{6sin(6 - 6cos2t)}{6 - 6cos2t}$
select the correct choice below and, if necessary, fill in the answer box.
a. $lim_{t
ightarrow0}\frac{6sin(6 - 6cos2t)}{6 - 6cos2t}=square$ (type an integer or a simplified fraction)
b. the limit does not exist.

Explanation:

Step1: Let $u = 6 - 6\cos2t$.

As $t
ightarrow0$, we have $u = 6-6\cos(0)=6 - 6\times1=0$.

Step2: Rewrite the limit.

The given limit $\lim_{t
ightarrow0}\frac{6\sin(6 - 6\cos2t)}{6 - 6\cos2t}$ can be rewritten as $6\lim_{u
ightarrow0}\frac{\sin u}{u}$ (since when $t
ightarrow0$, $u
ightarrow0$).

Step3: Apply the limit formula.

We know that $\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta}=1$. So, $6\lim_{u
ightarrow0}\frac{\sin u}{u}=6\times1 = 6$.

Answer:

A. $\lim_{t
ightarrow0}\frac{6\sin(6 - 6\cos2t)}{6 - 6\cos2t}=6$