Michael B. | I can provide your 'A-HA' momentI can provide your 'A-HA' moment
5.05.0(149 lesson ratings)(149)
There is a simple process for determining discontinuities and zeros of rational functions (that is, a ratio of two polynomial functions):
First, fully factor the numerator and denominator to the smallest order polynomials that have real coefficients. So you should factor x2-4 to (x+2)(x-2) but you don't factor x2+4 to (x+2i)(x-2i).
Then cancel any terms that are common in the numerator and denominator
Any terms that are left in the denominator create non-removable infinite discontinuities (that is, they do not have a limit) at those values of x that cause any of the remaining denominator terms to equal zero
Any terms that are left in the numerator create zeros (where the graph crosses the x-axis) at those values of x that cause the remaining numerator terms to equal zero
Excepting any infinite discontinuities determined in Step #3, any terms that were cancelled in Step 2 produce removeable discontinuities (that is, they do have a limit) for those values of x that cause the terms to equal zero. The limit value is the same as the function evaluated at that point after the offending terms have been cancelled out.
The (x-2) term cancels from numerator and denominator... Then, (x+3) is left in the denominator, so x = -3 is an infinite non-removeable discontinuity, and the limx->-3 does not exist. Then, x = -2 is a zero of the function, and x = 2 is a removeable hole. Therefore, limx->2 = (2+2)/(2+3) = 4/5. Had there been another (x-2) in the denominator in addition to the one that is there such as (x-2)2 then it would not be a hole, but would still be an infinite non-removeable discontinuity as specified in Step #3.
Helga M. | Lots of Experienc: 20 years College Teaching, 35 years TutoringLots of Experienc: 20 years College Teac...
to find if a discontinuous function has a limit, you will need to see how far you can simplify the function. In your case, you can factor the numerator as well to get