Question

How do you find the limit of #(x^2+x-6)/(x+3) as x approaches -3?

Answer 1

With quadratic terms, you factorise.

Explanation:

`=((x-2)(x+3))/((x+3))`

As long as `x!=-3` we may cancel:

`=((x-2)cancel((x+3)))/cancel((x+3))=x-2`

The limit for `x->-3` is `-5`, or:

`lim_(x->-3) f(x)=-5`

`(-3,-5)` is a discontinuity.
graph{(x^2+x-6)/(x+3) [-9.5, 10.5, -7.32, 2.68]}