# Limit of a function

1. Sep 10, 2013

### Ipos Manger

1. The problem statement, all variables and given/known data

Does lim x-> -2 (3x2+ax+a+3)/(x2+x-2) exist?

If so, find the limit.

2. Relevant equations

-

3. The attempt at a solution

I've tried factorizing the denominator to (x+2)(x-1), but then I don't know how to proceed on the exercise. I have seen that the limit exists when the numerator equals 0, but why is this (the denominator is also 0)? Does 0/0 exist at all?
Code (Text):

2. Sep 10, 2013

### Staff: Mentor

What value of a would make the numerator zero when x = -2?

When both numerator and denominator are zero (i.e., the indeterminate form 0/0), it's usually because there is a factor in common between the top and bottom. Although not certain, there's a good chance that a limit exists.

3. Sep 10, 2013

### Ipos Manger

a = 15.

So that means I just have to learn that by heart? Isn't there sort of a formal proof to test that? On the other hand, how do you know there's a factor in common between the top and bottom knowing that numerator and denominator are zero?

Thank you for your answer.

4. Sep 10, 2013

### Staff: Mentor

I don't think you need to memorize what I said - just be aware that if a limit has the form "0/0" you're not done yet. "0/0" is not a value - it is an indeterminate form, which means that you can't determine a value. All of the limits below have this form, but they have wildly different limit values.

$$\lim_{x \to 0}\frac {x^2}{x} = 0$$
$$\lim_{x \to 0}\frac {x}{x} = 1$$
$$\lim_{x \to 0^+}\frac {x}{x^2} = \infty$$

Note that in the 3rd limit, the two-sided limit doesn't exist, and that's the reason I am using the one-sided limit, the limit as x approaches zero from the right.

As far as a theorem goes, the only one that comes to mind is L'Hopital's Rule, which is applicable in cases with the indeterminate forms 0/0 and ±∞/∞.

5. Sep 11, 2013

### verty

After you have factored the denominator, factor the numerator (or use Mark44's shortcut).

6. Sep 12, 2013

### vela

Staff Emeritus
There's a theorem that says that if p(x) is a non-constant polynomial and p(a)=0, then p(x) can be factored into p(x)=(x-a)q(x), where q(x) is another polynomial. In this problem both the numerator and denominator are polynomials; therefore, if they both equal 0 at x=-2, they have (x-(-2)) as a common factor.

7. Sep 12, 2013

### chatnay

I'm sorry to hijack the thread, but does this mean that i have to learn as a general rule if the denominator is 0 (in the case of the limit), I have to solve the numerator for 0 and find the values that I can factorize? Thank you.

8. Sep 13, 2013

### verty

I would say no, you don't have to do that. You must try to factorize the numerator IF it goes to 0.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted