Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.3 One-sided Limits; Continuous Functions - 13.3 Assess Your Understanding - Page 910: 76

not continuous at $x=-4$ or $x=4$.

1. See graph for $R(x)=\frac{x^2+4x}{x^2-16}=\frac{x(x+4)}{(x+4)(x-4)}=\frac{x}{x-4}, x\ne-4$ with V.A. $x=4$, H.A. $y=1$ and a hole at $(-4, \frac{1}{2})$. 2. Although $\lim_{x\to-4}R(x)=\frac{1}{2}$, the function is not continuous at $x=-4$ because of the hole. 3. We have $\lim_{x\to 4^-}R(x)=-\infty$ and $\lim_{x\to 4^+}R(x)=\infty$, thus the function is not continuous at $x=4$.