## Precalculus (6th Edition) Blitzer

a. Using the given graph, we have $\lim_{t\to t_1}p(t)=20$ and $p(t_1)=20$ b. We can conclude that $p$ is continuous at $t_1$ because $\lim_{t\to t_1^-}p(t)=\lim_{t\to t_1^+}p(t)=p(t_1)$. That is, $p$ is defined at $t_1$ and the limit exists and is equal to the function value at $t_1$. c. Using the given graph, we have $\lim_{t\to t_2}p(t)=40$ and $p(t_2)=10$ d. We can conclude that $p$ is not continuous at $t_2$ because $\lim_{t\to t_2}p(t)\ne p(t_2)$ e. We have $\lim_{t\to t_3^-}p(t)=50$, $\lim_{t\to t_3^+}p(t)=70$, $p(t_3)=70$; thus $\lim_{t\to t_3}p(t)$ does not exist. f. We can conclude that $p$ is not continuous at $t_3$ because $\lim_{t\to t_3}p(t)$ does not exist. g. We have $\lim_{t\to t_4^-}p(t)=100$, $p(t_4)=100$, h. $t_4$ means the end of the course, and $100$ means the maximum percentage that can be achieved. The limit in (g) means that as time approaches the end of the course, the student learned close to $100\%$ of the content. The function value in (g) means that at the end of the course, the student learned $100\%$ of the content.