Chapter 14.3, Problem 40AYU

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

To determine

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

Answer to Problem 40AYU

2

Explanation of Solution

Given:

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

Calculation:

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

Then we have,

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