If i have rational function and it has a numerator that can be factored and the denominator is already factored out would I simplify by factoring the numerator?

Question

The question is:

((x^2)-4)/ ((x-2)(x+3))

and then cancelling out the x-2 from both the denominator and the numerator leaving me with (x+2)/(X+3)? And then go from there to find the limit?

Answer 1

Yes, you can simplify it that way, but then you need to note that your new Domain is all real numbers with the exception of 2.

Domain = R - {2} is how you would notate it.

 

Do you understand why? Do you know what makes two equations equilvent?

Answer 2

Yes,of course you can simplify it,but after simplifying the given rational equation( (×^2-4)/(x-2)(x+3)),it is important to note that the domain is the set of real number except 2.

As x approaches -3 from left 

Answer 3

I take it the question is to find the limit of the function as x approaches 2.

It doesn't matter whether (2, f(2)) is a point on a continuous function, or is an isolated point off a continuous function f, or doesn't exist at all. In any of those cases, the limit exists as long as the limit as x approaches 2 from the left exists, the limit as x approaches 2 from the right exists, and those two limits are the same. What's happening at 2 itself is irrelevant. And that's why it's OK to cancel the (x-2)'s! Because we only care about what's happening from the left and from the right of 2, not at 2 itself.

Answer 4

Yes, you ha e to cancel the common factor then plug X to find the limit. Since X- 2 is in the denominator so X -2 is not zero and X cannot be = 2 (X is approaching to 2) meaning the function is not defined at x=2 and it is not continuous at 2 so it has an empty hole in the graph.

Answer 5

I would actually add that simplifying (x+2) in the numerator and denominator is not related causally to the exclusion of 2 from the domain. Conditions of existence need be established before the equation (or function or ...) is manipulated. This ratio of polynomials exists only for x ≠ 2 and x≠ -3.

Existence should be the first step. After that's established, one can simplify without losing track of the various conditions and constraints.

Answer 6

The way you've simplified the function is correct. You divided out the "x-2" factor in both the numerator and denominator, and the limit you want would be the limit of (x+2)/(x+3). As others have said, keep in mind the domain does exclude 2; from a graphing standpoint there is a "hole" in the graph where you would ordinarily see (2, 0.8) in the "simplified" function.

Answer 7

Yes you have to factor the denominator or you won't know if simplification is possible at this stage.

Answer 8

Danielle,

To echo/amplify Jon's point: you can of course cancel (x-2), but keep in mind there is a conditon for you to be able to do that: x-2 /= 0; or in other words, the cancellation is only valid for x /= 2.