Precalculus: What is discontinuity in rational functions? Is it another way of saying there is a hole in the graph? Discontinuity: Anytime the denominator of a rational function equals zero. What if it's 0/0? Is it a discontinuity or a hole? Or is it both? Does it have to be a number over 0 to count as a discontinuity?
No, it is still a discontinuity. Just imagine the function f(x) = x/x does the point x=0 exist? Nope.
so 0/0 is not a hole?
It is a hole. The numerator does no have to be over 0 to count as a discontinuity.
How does it count as a discontinuity?
You cannot divide by zero, therefore the point does not exist
it is a jump discontinuity
Wait, isn't that what a vertical asymptote is?
Do you have a graphing calculator?
or just a regular calculator?
graphing calculator
Graph the function y = x/x then go to >2nd>calc>value x=0
It will give you an error because the point does not exist, it is 0/0. And it is a jump discontinuity. The condition for the discontinuity is that the denominator is 0, the numerator DOES NOT matter
You are confusing discontinuities with limits, where 0/0 could have an actual value (limit)
i guess you wanto knw how to knw whether a discontinuity has vertical asymptote or not
Wait, so in f(x) = 1/(3x^2+3x-18) The vertical asymptotes are 2,-3 which are also the discontinuities?
x=2 and x=-3
are vertical asymptotes and discontinuities the same thing?
vertical asymptotes are discontinuities but discontinuities are not necessarily vertical asymptotes
Understand?
Why can't discontiunities be vertical aympsotes?
sometimes the factors leading to 1/0 cancel out in numerator and denum (just the case with y=x/x) leaving a discontinuity(hole) without any asymptote, cuz around the hole, graph stays at fixed value 1 here, it doesnt increase forever...
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consider another example where factors leading to 1/0 doesnt cancel out. y = 1/x
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