# Does a limit exist?

1. Jul 6, 2012

### KiwiKid

1. The problem statement, all variables and given/known data
Is there a number 'a' such that

(3x^2 + ax + a + 3) / (x^2 + x - 2)

exists as x goes to -2? If so, find the value(s) of 'a' and the limit.

2. Relevant equations
The limit rules. Algebra.

3. The attempt at a solution
Well... I have absolutely no clue what I'm supposed to do. I suppose I'll have to choose my 'a' values in such a way so the denominator no longer equals zero, *but* other than brute-forcing a-integers in a graphical calculator I have little idea on how to do that.

I began by factoring the denominator to (x+2)(x-1), with the hope that I may be able to negate the x+2 somehow, but that didn't work. Then I tried to get (x^2+x-2) in the numerator so the numerator and denominator would cancel out and leave a bit on top, but I couldn't find a way to do that, either.

Could someone give me a clue on what approach to try here?

2. Jul 6, 2012

### micromass

Staff Emeritus
The denominator always goes to 0 if x goes to -2. What should the numerator go to if the limit were defined?? For example, if the numerator were 1, then the limit would be "1/0" which would be an infinity and thus the limit would not exist.

Is there a value for the numerator for which the limit does exist?

3. Jul 6, 2012

### Whovian

Have you heard of L'Hopital's Rule? It should work here. If not, note that the numerator must also go to 0 as x goes to -2, and must therefore have a factor of x+2.

4. Jul 6, 2012

### KiwiKid

Are you saying that the numerator should equal 0?

5. Jul 6, 2012

### micromass

Staff Emeritus
Yes! If x=-2, then the numerator should be 0.

6. Jul 6, 2012

### KiwiKid

Ok, let me see if I got this right:

3(-2)^2+a(-2)+a+3 = 0
-> 12 - 2a + a + 3 = 0
-> a = 15

Therefore (substituting a):
lim[x->-2] (3x^2 + 15x + 18) / (x^2 + x - 2)
= lim[x->-2] (3(x+2)(x+3)) / ((x+2)(x-1))
= lim[x->-2] (3(x+3)) / (x-1)
= 3 / -3
= -1

I think I got it. Thanks again!

7. Jul 6, 2012

### SammyS

Staff Emeritus
That looks good.