Question

In Problems 33– 44, find the one-sided limit.

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

Short answer

The value of limit is

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

Step-by-step solution

step 1.introduction

Given limit

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

step 2.equation simplification

consider the equation

$$\begin{gathered} =\frac{x^2-1}{x^3+1} \\ =\frac{(x-1)(x+1)}{(x+1)(x^2-x+1)} \\ =\frac{(x-1)}{(x^2-x+1)} \end{gathered}$$

step 3.the solution

the limit value

$$\begin{gathered} \underset{x\to -1^{-}}{\lim \frac{x^2-1}{x^3+1}}=\underset{x\to -1^{-}}{\lim \frac{x-1}{x^2-x+1}} \\ =\frac{-1-1}{{(-1)}^2-(-1)+1} \\ =\frac{-2}{3} \end{gathered}$$