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: 80

hole at $x=1$, asymptote at $x=-4$.

1. Factor to get $R(x)=\frac{x^3-x^2+3x-3}{x^2+3x-4}=\frac{x^2(x-1)+3(x-1)}{(x+4)(x-1)}=\frac{(x-1)(x^2+3)}{(x+4)(x-1)}=\frac{x^2+3}{x+4}, x\ne1$, thus the rational function is undefined at $x=1$ and $x=-4$. 2. We can determine a hole at $x=1$ and an asymptote at $x=-4$.