Question

use the relation $lim_{\theta
ightarrow0}\frac{sin\theta}{\theta}=1$ to determine the limit $lim_{t
ightarrow0}\frac{6sin(4 - 4cos7t)}{4 - 4cos7t}$

Explanation:

Step1: Let $\theta = 4 - 4\cos7t$

As $t
ightarrow0$, we find the limit of $\theta$. We know that $\lim_{t
ightarrow0}\cos7t=\cos(0) = 1$. So, $\lim_{t
ightarrow0}(4 - 4\cos7t)=4-4\times1 = 0$.

Step2: Rewrite the given limit

We are given $\lim_{t
ightarrow0}\frac{6\sin(4 - 4\cos7t)}{4 - 4\cos7t}$. We can rewrite it as $6\times\lim_{t
ightarrow0}\frac{\sin(4 - 4\cos7t)}{4 - 4\cos7t}$.

Step3: Apply the limit formula

Since $\lim_{\theta
ightarrow0}\frac{\sin\theta}{\theta}=1$ and as $t
ightarrow0$, $4 - 4\cos7t
ightarrow0$, we have $6\times\lim_{t
ightarrow0}\frac{\sin(4 - 4\cos7t)}{4 - 4\cos7t}=6\times1$.

Answer:

$6$