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))

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))

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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?

Amaan M.

Math/Economics Teacher

New York, NY

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John P.

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## Comments

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) /(x

^{2}- 5x – 6)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?

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.

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?