Chapter 14.2, Problem 38AYU

To determine

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

Answer to Problem 38AYU

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

Explanation of Solution

Given:

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

Calculation:

Simplify the expression as follows.

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

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