Michael B. answered • 11/13/12

I can provide your 'A-HA' moment

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 x
^{2}-4 to (x+2)(x-2) but you don't factor x^{2}+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.

For your example:

` x`

^{2} - 4 (x-2)(x+2)

---------- = ----------

(x-2)(x+3) (x-2)(x+3)

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 lim_{x->-3} does not exist. Then, x = -2 is a zero of the function, and x = 2 is a removeable hole. Therefore, lim_{x->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.

I believe you mistyped. The limit exists at x = 2, not x=-2. :-)

10/01/12