Danielle P.

# how do i find where a function is discontinuous if the bottom part of the function has been factored out?

my problem is:

Indicate all values for which the function is discontinuous. for each discontinuity point "a" give the limit of the function as x approaches "a". Clearly indicate limits that do not exist.

(10-x)/ ((x-2)(x-3))

By:

Mathematics/writing tutor, with degrees in math and education

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?

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09/30/12

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.

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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?

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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?
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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)

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09/30/12

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