Answer
$\lim\limits_{t\to0}\dfrac{\sin{3t}}{2}=\dfrac{3}{2}.$
Work Step by Step
$\lim\limits_{t\to0}\dfrac{\sin{3t}}{2t}=\frac{3}{2}\lim\limits_{t\to0}\dfrac{3\sin{3t}}{3t}=\frac{3}{2}\lim\limits_{t\to0}\dfrac{\sin{3t}}{3t}.$
Let $y=3t.$ Notice that as $t\to0$; $y\to0$
$\frac{3}{2}\lim\limits_{y\to0}\dfrac{\sin{y}}{y}=(\frac{3}{2})(1)=\dfrac{3}{2}.$
