## Thomas' Calculus 13th Edition

$\dfrac{3}{4}$
We know $$\lim\limits_{\theta \to 0} \dfrac{\mathrm{sin}\left(\theta \right)}{\theta }=1.$$ Thus for any nonzero real number $a$, it follows that $$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=1.$$ We will use this fact to compute the limit below. So to see that this is true, we let $x=a\theta.$ Then since $x \to 0$ as $\theta \to 0$, we have $$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=\underset{x\to 0}{\lim}\dfrac{\mathrm{sin}\left(x\right)}{x}=1.$$ Now, we want to find $\underset{y\to 0}{\lim}\dfrac{\mathrm{sin}\left(3y\right)}{4y}$. Note that $$\dfrac{\mathrm{sin}\left(3y\right)}{4y}=\dfrac{3\mathrm{sin}\left(3y\right)}{3\left(4y\right)}=\dfrac{3}{4}\dfrac{\mathrm{sin}\left(3y\right)}{3y}.$$ Thus $$\underset{y\to 0}{\lim}\dfrac{\mathrm{sin}\left(3y\right)}{4y}=\underset{y\to 0}{\lim}\left(\dfrac{3}{4}\frac{\mathrm{sin}\left(3y\right)}{3y}\right)=\dfrac{3}{4}\left(\underset{y\to 0}{\lim}\dfrac{\mathrm{sin}\left(3y\right)}{3y}\right)=\dfrac{3}{4}\left(1\right)=\dfrac{3}{4}.$$