Question

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

$\underset{x\to 2}{\lim }\frac{x^3-2x^2+4x-8}{x^2+x-6}$

Short answer

The limit rounded to two decimal places is$1.6$

Step-by-step solution

Step 1. Given information

The given limit is

$\underset{x\to 2}{\lim }\frac{x^3-2x^2+4x-8}{x^2+x-6}$

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 for the given limit is:

Therefore$\underset{x\to 2}{\lim }\frac{x^3-2x^2+4x-8}{x^2+x-6}=1.6$