Chapter 14.2, Problem 40AYU

To determine

To find: The value of limit $\underset{x\to 1}{\lim }\frac{x^3-x^2+3x-3}{x^2+3x-4}$ .

Answer to Problem 40AYU

The value of limit $\underset{x\to 1}{\lim }\frac{x^3-x^2+3x-3}{x^2+3x-4}$ is $\frac{4}{5}$ .

Explanation of Solution

Given:

The function is $\underset{x\to 1}{\lim }\frac{x^3-x^2+3x-3}{x^2+3x-4}$ .

Calculation:

Simplify the expression as follows.

  $\begin{array}{c}\underset{x\to 1}{\lim }\frac{x^3-x^2+3x-3}{x^2+3x-4}=\underset{x\to 1}{\lim }\frac{x^2\left(x-1\right)+3\left(x-1\right)}{\left(x-1\right)\left(x+4\right)}\\ =\underset{x\to 1}{\lim }\frac{\left(x-1\right)\left(x^2+3\right)}{\left(x-1\right)\left(x+4\right)}\\ =\frac{1^2+3}{1+4}\\ =\frac{4}{5}\end{array}$

Hence, the value of limit $\underset{x\to 1}{\lim }\frac{x^3-x^2+3x-3}{x^2+3x-4}$ is $\frac{4}{5}$ .