Expert: Paul Klarreich - 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
Questioner:   Steph
Category:  Advanced Math
Private:  No
 
Subject:  Limit Dilemma
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.
............................................
Hi, Steph,

All of them.  Of course, it depends on what you want to do.  I conclude from the date (it is late September, when young calculus students are just learning to do limits) that you are a young calculus student just learning to do limits.

So here is what you are doing:

1. Try  x = -3.  You get  24/0.  Bad news.  f(-3) does not exist.  But  lim(x->-3) f(x) could still exist.  lim(x->0) is based on values of x that are NOT -3.

2. Try to simplify.  (x+3) will cancel out (as you noted), and canceling (x+3) is legal PROVIDED x /= -3.  And, as we noted, lim(x->-3) is based on  x /= -3.

3. Apply  x = -3 to the simplified function.  The mathematical principle working here is that:

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

are the same function, PROVIDED  x /= -3.

AND f1(x) = x - 4  is continuous, so  lim(x->-3) f1(x) = f1(-3) = -7.

A few hundred more of these and you'll feel more comfortable with it.

So, basically, you do #1, and if necessary, #3.  Save #2 for later, when you do curve sketching, and write me again.

[No, no, you don't have to wait that long.]