Chapter 14.3, Problem 41AYU

$\underset{x\to -1^{-}}{\lim }\frac{x^2-1}{x^3+1}$

To determine

To find: One sided limit: $\underset{x\to -1^{-}}{\lim }\left(\frac{x^2-1}{x^3+1}\right)$.

Answer to Problem 41AYU

$-\frac{2}{3}$

Explanation of Solution

Given:

$\underset{x\to -1^{-}}{\lim }\left(\frac{x^2-1}{x^3+1}\right)$

Calculation:

Since $\underset{x\to -1^{-}}{\text{lim}}$; so $x\ne -1$ or $x+1\ne 0$.

Then we have,

$\underset{x\to -1^{-}}{\lim }\left(\frac{x^2-1}{x^3+1}\right)=\underset{x\to -1^{-}}{\lim }\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\underset{x\to -1^{-}}{\lim }\frac{\left(x-1\right)}{\left(x^2-x+1\right)}=\frac{\left(-1-1\right)}{\left[{\left(-1\right)}^2-\left(-1\right)+1\right]}=\frac{\left(-2\right)}{\left(1+1+1\right)}=-\frac{2}{3}$