Chapter 14.3, Problem 43AYU

To determine

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

Answer to Problem 43AYU

$\frac{3}{2}$

Explanation of Solution

Given:

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

Calculation:

Since $\underset{x\to -2^{+}}{\text{lim}}$; so $x\ne -2$ or $x+2\ne 0$.

Then we have,

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