Chapter 14.2, Problem 41AYU

To determine

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

Answer to Problem 41AYU

The value of limit $\underset{x\to -1}{\lim }\frac{x^3+2x^2+x}{x^4+x^3+2x+2}$ is 0.

Explanation of Solution

Given:

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

Calculation:

Simplify the expression as follows.

  $\begin{array}{c}\underset{x\to -1}{\lim }\frac{x^3+2x^2+x}{x^4+x^3+2x+2}=\underset{x\to -1}{\lim }\frac{\left(x+1\right)\left(x^2+x\right)}{\left(x+1\right)\left(x^3+2\right)}\\ =\underset{x\to -1}{\lim }\frac{x^2+x}{x^3+2}\\ =\underset{x\to -1}{\lim }\frac{{\left(-1\right)}^2-1}{{\left(-1\right)}^3+2}\\ =0\end{array}$

Hence, the value of limit $\underset{x\to -1}{\lim }\frac{x^3+2x^2+x}{x^4+x^3+2x+2}$ is 0.