# Chapter 1 - Limits and Their Properties - 1.4 Exercises: 48

$f(x)$ has a removable discontinuity at $x=-2$ and a non-removable discontinuity at $x=3.$

#### Work Step by Step

$f(x)=\dfrac{x+2}{x^2-x-6}=\dfrac{(x+2)}{(x+2)(x-3)}=\dfrac{1}{x-3}\to x\ne-2$ and $x\ne3.$ Hence, $f(x)$ has a removable discontinuity at $x=-2$ and an irremovable discontinuity (vertical asymptote) at $x=3.$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.