$limit and discontinous$

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$limit and discontinous$

Can someone give me an example of two functions whose domain is R^2 and codomain R^2, the first is discontinuous function at the point (1,2), and the second one has no limit at (1,2), and for both functions it is necessary to prove epsilon delta?

Multivariable Calculus Limits Functions

Jivs

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You should formulate your questions more carefully and propose a "bounty" in adequation of what you expect... But here's a valid answer:

You can use the function f(x,y) = (c(x),c(y)) where c is the characteristic function of irrationals: c(x)=0 for rational x (i.e., x = p/q, with integers p and q>0) and c(x)=1 for irrational x. (You can also take te second component equal to c(x), or even equal to zero.)

The function works for both cases: it's disconinuous and has no limit at (1,2) where it equals o = (0,0) [since 1 & 2 are rational]:

Indeed, for any $\delta>0$, as small as you wish, you always have points (x,y) with irrational x and/or y such that $\|(x,y)- (1,2)\| < \delta $:
You can take $x=1+\sqrt2/N$ and $y=x+1$, then $\|(x,y)- (1,2)\|=2/N < \delta$ for any integer $N>2/\delta$.
At these points, the function has at least one component equal to 1, and therefore $ \| f(x,y) - f(1,2) \| \ge 1 > \epsilon $ whenever $0 < \epsilon < 1$.
So it does not have a limit in (1,2) which also means that it is discontinuous at (1,2).

"It is necessary to prove epsilon delta" doesn't mean anything, but the above is indeed the proof of discontinuity using $\epsilon$ and $\delta$.

PS: if you really want two distinct functions, take f(x,y)=(c(x),0) and g(x,y) = (0,c(x)). Then f(x,y) = (1,0) and g(x,y) = (0,1) whenever x is irrational, so $\| f(x,y) - f(1,2) \| = \| g(x,y) - g(1,2) \| = 1$ as for the example given above.

M F H

356

Jivs

+1

Thank you very much
- M F H
  
  0
  
  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. (
Jivs

0

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?
- M F H
  
  0
  
  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.
M F H

0

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.
- M F H
  
  0
  
  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...]
M F H

0

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).
Jivs

0

okay thank you

$limit and discontinous$

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