Answer
a. $1$
b. $\frac{2}{3}$
Work Step by Step
a. As $\theta\to 3^+$, $\lfloor\theta\rfloor\to3$, we have $\lim_{\theta\to3^+}\frac{\lfloor\theta\rfloor}{\theta}=\frac{3}{3}=1$
b. As $\theta\to 3^-$, $\lfloor\theta\rfloor\to2$, we have $\lim_{\theta\to3^-}\frac{\lfloor\theta\rfloor}{\theta}=\frac{2}{3}$
