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: 82

hole at $x=3$, asymptote at $x=-1$.

1. Factor to get $R(x)=\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}=\frac{x^2(x-3)+4(x-3)}{x^3(x-3)+(x-3)}=\frac{(x-3)(x^2+4)}{(x-3)(x^3+1)}=\frac{x^2+4}{x^3+1}, x\ne3$, thus the rational function is undefined at $x=3$ and $x=-1$. 2. We can determine a hole at $x=3$ and an asymptote at $x=-1$.