Question

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

$\underset{x\to -4^{-}}{\lim }\frac{x^2+x-12}{x^2+4x}$

Short answer

The value is$\frac{7}{4}$

Step-by-step solution

Given information

Given function$\underset{x\to -4^{-}}{\lim }\frac{x^2+x-12}{x^2+4x}$

Sketching the graph

The graph is

Checking the limit

Checking, we get

For $x\to 4-$,$x$is close to four but less than $4$so

$\underset{x\to -4^{-}}{\lim }\left(\frac{x^2+x-12}{x^2+4x}\right)=\frac{7}{4}$