limit and discontinous
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Thank you very much
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You're welcome. I made some improvements. Also, instead of the characteristic function you could also use c(x)=sin(1/(x-1)) for nonzero 0 and c(0)=0, or any other function that is discontinuous at x=1. (
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thank you, you are amazing. If I would use c(x)=sin(1/(x-1)), I could define function f(x,y)=(sin(1/(x-1)) ,3) if (x,y)!=1,2 and f(x,y)=(0,0) if (x,y)=1,2 and that would be good example?
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No, there's a little problem, you can be in a point different from (1,2) but with x=1 where sin(1/(x-1)) isn't defined. So you should rather say: the first if x != 1, and the second if x = 1.
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But if you are required to make the epsilon-delta proof then, it depends whether it will be accepted if you just say "sin(1/(x-1)) doesn't have a limit" or if you have to exhibit points (with |x-1| x = 1+ 2/(N pi), with N large enough so that |x-1| < delta, and then if N is odd this is +-1 and if N is even this is zero.
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oops, my comment got scrambled ... something is lost between "|x-1|" and "x = 1+2/(N pi)". It should read: "...(with |x-1| < delta), such that sin(...) = +-1 or 0. I.e., 1/(x-1) = N pi/2 , then < = > x = 1+2/(N pi) ... [I think it's the < sign that makes an HTML tag that hides my text... Trying to insert spaces around...]
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If you let f(x,y)=(0,0) whenever x=1 then you do have a limit along that line {x=1}, but you don't have a limit as (x,y) -> (1,2) from any other direction, so your function still isn't continuous and the limit doesn't exist. (again, because there are point closer than any delta>0 in which the function is "far from (0,0)" and actually also far from any other point, since it goes through (1,3) and through (-1,3) and through (0,3) infinitely often in any neighborhood of (1,2).
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okay thank you
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