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: 77

hole at $x=1$, asymptote at $x=-\sqrt[3] 2$.

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