Find Limit of Function: Does 0/0 Exist?

[Ipos Manger]

Homework Statement

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

If so, find the limit.

Homework Equations

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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?

 

[Mark44]

Homework Statement

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

If so, find the limit.

Homework Equations

-

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?

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.

 

[Ipos Manger]

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.

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.

 

[Mark44]

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 ±∞/∞.

 

[verty]

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

 

[vela]

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?

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.

 

[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.

 

[verty]

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.

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