Chapter 2 - Limits - 2.6 Trigonometric Limits - Exercises - Page 77: 19

$3$

\begin{align*} \lim _{t \rightarrow 0} \frac{\sqrt{t^3+9}\sin t}{t}&=\lim _{t \rightarrow 0} \frac{\sin t}{t}\sqrt{t^3+9}\\ &=\lim _{t \rightarrow 0} \frac{\sin t}{t}\lim _{t \rightarrow 0}\sqrt{t^3+9}\\ &=3. \end{align*}