Expert: Steve Holleran - 9/25/2008

Question
Hi
I have a limit that looks as such:
lim x -> -3
(x^2-x+12) / (x+3).
1. When I sub in the -3 for x, I get the limit as 24/0.
2. When I redo the limit as one sided, I get a + infinity / - infinity, meaning the limit does not exist.
3. If I factor out the equation, the (x + 3 ) will cancel out and I get an answer of -7.
Which of these 3 methods is the appropriate and correct method?
Thank you.

Answer
Hi Steph,

I think you were initially correct, and may have a factoring mistake in your third point.

On limits, you ALWAYS want to substitute first:

 lim (x-> -3) [ x^2 - x + 12 / x + 3]  =  24 / 0

This means there is some infinite behavior going on, and the limit definitely does not exist.

If you try to factor the numerator , x^2 - x + 12,
you would need factors of 12 that add to -1, and there aren't any.
I think you may have factored it as

           x^2 - x + 12 = (x - 4)(x + 3)

but that would be x^2 - x - 12.

So, you're first inclination was correct.

Steve