Question

Use a graphing utility to find the indicated limit rounded to two decimal places.

$\underset{x\to 3}{\lim }\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}$

Short answer

The value of the limit rounded to two decimal places is$0.46$

Step-by-step solution

Step 1. Given information

The given limit is$\underset{x\to 3}{\lim }\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}$

Using a graphing utility we will find the table of values to find the value of the given limit.

Step 2. Table of values

The table of values is:

Therefore $\underset{x\to 3}{\lim }\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}=0.46$