Chapter 14 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - 14.1 Finding Limits Using Tables and Graphs - 14.1 Assess Your Understanding - Page 877: 48

$\approx 0.46$

We have to determine the limit: $\displaystyle\lim_{x\rightarrow 3} \dfrac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}$ We use a graphing utility to compute the limit. The result is: $\displaystyle\lim_{x\rightarrow 3} \dfrac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}\approx 0.46$