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Homework Help: Trouble understanding Continuity

  1. Jun 8, 2008 #1
    Hey folks,
    I have this one problem that seems unclear to me:

    Show that the function f is continuous at (0,0) for
    f(x,y) = ysin(1/x) if (x,y) do not equal (0,0)..........and 0 if (x,y) = (0,0)

    I'm thinking though, as with single variable calc, f is continuous at (a,b) provided that f(a,b) exists right? Therefore the top function is not continuous at (0,0), correct? But then again it states that x and y never do equal (0,0). I think I don't understand the question. Could somebody rephrase it or give me a hint? Cheers
  2. jcsd
  3. Jun 8, 2008 #2
    I think it would be good for you to start with the equivalent 1 dimensional problem, which is:

    Let f(x) = x sin(1/x) for all x not equal to zero, and define f(0) = 0. Show that f is continuous everywhere.
  4. Jun 8, 2008 #3


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    No, even in single variables that is not right. f is continuous at (a,b) provided that (1) f(a,b) exists, (2) [itex]lim_{(x,y)\rightarrow (a,b)}[/itex] exists, and (3)[itex]lim_{(x,y)\rightarrow (a,b)}= f(a,b)[/itex].

    Even from your mistaken statement that does not follow. f(0,0) certainly does exist. You are told that "f(x,y)= 0 if (x,y)= (0,0)". In other words that f(0,0)= 0.
    Your reference to "the top function" implies that you think there are two functions here. That is incorrect. There is only one function.

    No, it does NOT say "x and y never do equal (0,0)". It says that if x and y are not (0,0) then f(x,y) has a certain formula.

    Suppose a function f(x) were defined by "f(x)= 2x- 3 if x is not 1, 2 if x is 1", what is f(1.001)? What is f(1)? Is that function continuous at x= 1?
    Last edited by a moderator: Jun 8, 2008
  5. Jun 8, 2008 #4
    Thanks for the helpful insight so far.
    It is now clear to me that f(0,0) exists and is zero, so basically all I have to do to show that the function is continuous is prove that the limit of ysin(1/x) exists and is zero as (x,y) goes to (0,0).
    Now, how is that possible if the limit of (1/x) does not exist as x goes to zero? Is l'Hopitals rule somehow applicable in this situation?
  6. Jun 8, 2008 #5


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    No one has said that lim(1/x) exists as x goes to 0!

    What is being said is that (in your problem lim ysin(1/x) exists as x and y both go to 0 (independently) and (Crosson's one dimension example) x sin(1/x) exists as x goes to 0. What happens of x sin(1/x) for x very small? How large is "x"? How large is sin(1/x)? How large is their product?
  7. Jun 8, 2008 #6


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    Use the epsilon-delta definition of limit and the fact that |sin(z)| <= 1 for all z. It will be clear that delta=epsilon does the trick.
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