Locke

can you have a point be defined and undefined at the same time?
Ex: limit as x approaches 2 from the right f(x)=1 and f(2)=1. They're on the same graph.

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CSflim

a function where f(x) =f(y)=z ; x != y is perfectly valid. (It just means that the function is not injective)
But I don't see anything about a point being defined and undefined at the same time?

If you post more detail perhaps we could help you out more?

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Peetster

There is a difference between "limit as x approaches" and "x=". When x approaches a value, it never equals the value. It just gets really, really close.

KnifeMissile

Your question isn't clear.
What do you mean by point?
What do you mean by defined?

if f(x) = 1/x, then the function is undefined at x = 0 because it's definition cannot be evaluated at that point on the real line. The point (x,f(x)) on the Cartesion plane also has no meaning for x = 0, so perhaps you can say that that point is undefined but I suspect this isn't what you're thinking...

The example you've given doesn't suggest that the point at (2,1) is undefined--on the contrary, it is very well defined 'cause you defined it to be (2,1)!

Not only that but the function you've describe (as little as you described it) isn't even particularly interesting. Perhaps it was not your intention but your function is still continuous. What you said was:

lim(x->2)f(x) = 1 and f(2) = 1

Perhaps you meant to say something like this:

lim(x->2)f(x) = 3 and f(2) = 1

...? Even in this case, while the function is discontinuous, it is still well defined at x = 2.

Locke

Oh duh, It never touches that point it jst gets really close. Forgot about that. It works then. Thanks for the help.

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CUBS WIN, CUBS WIN!!!! - Pat Hughes
"Don't surround yourself with yourself." Yes