Answer
$$\lim_{y\to0}\frac{\sin 3y}{4y}=\frac{3}{4}$$
Work Step by Step
$$A=\lim_{y\to0}\frac{\sin 3y}{4y}$$
To be able to apply Theorem 7, we need $3y$ in the denominator, instead of $4y$.
So we would multiply both numerator and denominator by $3/4$:
$$A=\lim_{y\to0}\frac{3/4\sin3y}{3/4\times4y}=\frac{3}{4}\lim_{y\to0}\frac{\sin 3y}{3y}$$
Take $\theta=3y$. Then as $y\to0$, $\theta\to3\times 0=0$
$$A=\frac{3}{4}\lim_{\theta\to0}\frac{\sin\theta}{\theta}$$
Apply Theorem 7: $$A=\frac{3}{4}\times1=\frac{3}{4}$$
