removable discontinuity or non? f(x) = x(x^2 - x)

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kpx001

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how do i figure out the nonremobable and removable discontinuity for the function
f(x) = x/ (x^2-x) ? i got the nonremovable by factoring x/x(x-1) = 1/(x-1) and x= 1 , but i dunno how to get the removable discontinuity. i also no its an asymptote and im trying to figure this out anallyticaly
 
Re: discontinuity

kpx001 said:
how do i figure out the nonremobable and removable discontinuity for the function
f(x) = x/ (x^2-x) ? i got the nonremovable by factoring x/x(x-1) = 1/(x-1) and x= 1 , but i dunno how to get the removable discontinuity. i also no its an asymptote and im trying to figure this out anallyticaly
Click to expand...

For your given function x = 0 is the removable discontinuity. Look up the definition of removable discontinuity in your textbook - or Google it.
 
I once had to explain this very thing for a newspaper article. The reporter wanted to understand what the thing was in the background of a portrait he had taken for publishing. It was a sculpture of a nonremovable discontinuity.

After a few attempts, he seemed to understand:

Discontinuity - Broken
Removable - Fixed with a single point.
NonRemovable - Any repair more substantial than a single point.

Not exactly graduate level mathematics. :wink:
 
is there a way i can work out the solution to find what can be replaced? its a bit confusing because i expected since a limit would have a hole, that hole needs to be filled with the same x value.
 
You're a little backwards. Think about the Domain.

For \(\displaystyle f(x)\;=\;\frac{x}{x(x-1)}\), the Domain is all Real Numbers EXCEPT x = 1 and x = 0.

After you play with it, \(\displaystyle g(x)\;=\;\frac{1}{(x-1)}\), the Domain is all Real Numbers EXCEPT x = 1.

The really big question is, what happened to the Domain? What is different? Is f(x) EXACTLY THE SAME as g(x)? No! What is different? That is where the hole is.

Both f(x) and g(x) have an asymptote at x = 1. That isn't a hole.
 
ok i understand that now, the final thing in my mind is how to find the limit hole without being given a lim->c
 
Magic trick: g(x) is continuous. Use it to solve your dilemma.
 
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