Jonathan F. answered • 09/30/12
Mathematics/writing tutor, with degrees in math and education
Hi Danielle.
First you need to know where rational functions (do you know what a rational function is?) are discontinuous. Rational functions are discontinuous at any value where the denominator would be 0.
Next ask yourself where is the denominator 0 in your equation? In other words where is (x-2)(x-3) = 0
Do you know how to deal with the limits?
Jonathan F.
You can have a limit at a point outside of the domain. So the limit as x approaches 2 would be 4/5 (as you saw from your simplification), but the point 2 is discontinuous and not actually a part of the function. If you were to graph it you would make an open circle on the graph where x=2 to represent this.
The limit as x approached -3 would be -8, you solve that one normally.
09/30/12
Jonathan F.
Sorry it pasted funky when i tried to help you with -3.
As x approaches -3 from the left does it go to + or - inf? When it approaches from the left does it go to + or - inf?
09/30/12
Danielle P.
Thank you that makes sense! One more thought... If i have rational function (as you call it)and it has a numerator that can be factored (for example: the numerator is (x^2)-4) and the denominator is already factored out to be (x-2)(x+3), would I simplify it first by factoring the numerator 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?09/30/12
Jonathan F.
That's a great question!
Ask yourself this are fractions in the form of 0/0 allowed in math? The answer: not at all. You just can't factor a 0 out of the denominator. So in your example the function would still be discontinuous at x=-2.
Now here is a truly interesting question (one your class should be answering in a few days), what would the limit look like as it approached -2? Would it be different than the problem you just worked with?
I do think you have a slight misunderstanding though. The fact that the denominator is factored doesn’t impact the answer in any way. (10-x)/ ((x-2)(x-3)) is discontinuous at 2 and 3, but so is (10-x) /(x2 - 5x – 6)
09/30/12
ZOREH S.
12/19/21
Danielle P.
okay so the rational function ((x^2)-4)/((x-2)(x+3)) i just described would still be discontinuous at x=2 and x=-3, correct?
and then when it is time to find the limit for both of them there limits would not exist because the limits are -infinity and infinity because as x approaches both numbers the function gets bigger and bigger? or am i totally off base?
09/30/12