Answer
$$\frac{1}{2}$$
Work Step by Step
\begin{align*}
\lim _{t\rightarrow 0} \frac{\csc 8t}{\csc 4t} &=\lim _{t\rightarrow 0} \frac{\sin 4t}{\sin 8t} \\
&= \lim _{t\rightarrow 0} \frac{4}{8} \frac{\sin 4t}{4t} \frac{8t}{\sin 8t} \\
&= \frac{4}{8}\lim _{4t\rightarrow 0} \frac{\sin 4t}{4t} \lim _{8t\rightarrow 0}\frac{8t}{\sin 8t} \\
&=\frac{4}{8}=\frac{1}{2}.
\end{align*}
Where we used Theorem 2 -- that is, $\lim _{x\rightarrow 0}\frac{ \sin x}{ x}=1. $
