## University Calculus: Early Transcendentals (3rd Edition)

(a) $\lim_{t\to4^+}(t-[t])=0$ (b) $\lim_{t\to4^-}(t-[t])=1$
The graph is shown below. (a) $$\lim_{t\to4^+}(t-[t])$$ Looking at the graph, we see that as $t$ approaches $4$ from the right, $f(t)=t-[t]$ approaches $0$. Therefore, $$\lim_{t\to4^+}(t-[t])=0$$ (b) $$\lim_{t\to4^-}(t-[t])$$ Looking at the graph, we see that as $t$ approaches $4$ from the left, $f(t)=t-[t]$ approaches $1$. Therefore, $$\lim_{t\to4^-}(t-[t])=1$$