1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving Continuity Everywhere

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data

    Define
    f = { [itex]x^2[/itex] if [itex]x \geq 0[/itex]
    [itex]x[/itex] if [itex]x < 0[/itex]​

    At what points is the function f | [itex]\Re -> \Re[/itex] continous? Justify your answer.

    2. Relevant equations

    A function f from D to R is continuous at x0 in D provided that whenever {xn} is a sequence in D that converges to x0, the image sequence {f(xn)} converges to f(x0).

    3. The attempt at a solution

    Check x=0. {1/n} converges to 0. {f(1/n)} converges to 1. f(0) = 0. 0 ≠ 1. Thus the function is not continuous at x=0.

    When it comes to proving it's continuous elsewhere...that's where I have a problem.

    The example in the book for proving that a function is continuous at every other point simply states that "If a sequence {xn} converges to x0, then there is an index N such that f(xn) = f(x0) for all indices n > N.
    Thus, lim (n->∞) f(xn) = f(x0), and so f is continuous at the point x0."

    How is that a proof? That logic doesn't even exclude x=0. Can someone either elucidate what they've done, or explain an alternate way of proving this function's continuity at all points other than x=0?

    Thanks in advance.
     
  2. jcsd
  3. Oct 1, 2011 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    f(1/n) does not converge to 1. It converges to zero.
     
  4. Oct 1, 2011 #3

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    As SammyS points out {1/n} → 0. But that doesn't settle the question anyway. You have to show the limit works for all sequences converging to 0 if you're going to prove it that way.
    No, I don't believe your book says that.

    Do you have theorems such as "polynomials are continuous"? You can use that to show continuity for x away from 0. At 0 you need to show the right and left limits are both f(0).
     
  5. Oct 2, 2011 #4
    Sorry, I copied the problem incorrectly. The piecewise function should read:

    f(x) = x^2, if x<= 0
    x+1, if x>0​

    Thus, again:

    Check x=0. {1/n} converges to 0. {f(1/n)} converges to 1. f(0) = 0. 0 ≠ 1. Thus the function is not continuous at x=0.

    This would be correct if I was trying to prove that it is continuous at x=0, but I'm showing that it is NOT continuous at x=0. Thus a counterexample is sufficient. What I'm having trouble with is discovering a general method of showing that ALL of a function is continuous (excluding the point x=0)

    My text is Fitzpatrick Advanced Calculus 2e, and I've scanned the page in question.

    bogusproof.jpg

    No such theorem, at least not for a few sections. I think I understand, though. Using the sum and product properties of limits, I can say that for x<0, lim (n->∞) f(xn) = lim (n->∞) (xn2) = (lim (n->∞) x0))*(lim (n->∞) x0) = x02 = f(x0)

    Then after repeating the same process for x>0, I would conclude that f(x) is continuous from (-∞,0)U(0,∞).

    I suppose this is what that particular example in the book was trying to show, but it was woefully inadequate as an illustration of the process and seemed more general than it was meant to. I don't much like this textbook.
     
    Last edited: Oct 2, 2011
  6. Oct 2, 2011 #5

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The only reason that quote is correct is because the specific example to which it refers is constant nearby x if x ≠ 0. Very misleading to quote that "theorem" out of context.
     
  7. Oct 2, 2011 #6
    I never claimed it was a theorem, and indeed referenced it as part of an example. But, yes, I can see where it was misleading.

    My question stands, though. How is that sufficient justification, even within the example I've scanned? It seems to be glossing over important steps such as establishing a neighborhood of x for which the limit holds. Not to mention that my text doesn't establish the definition of a neighborhood until the next chapter.

    Such a lax progression of "logic" seems out of place in what is supposed to be a formal proof, especially when it is the sole example the text offers order for proving continuity in an interval.

    What I am asking is if someone can more clearly delineate those last few steps, because I do not feel this example sufficient for my understanding.

    Thanks.
     
  8. Oct 2, 2011 #7

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I tend to agree with you that it isn't the best example and it isn't written in complete detail. It depends on how much detail is required. For example, he just asserts 1/n → 0 as n → ∞ without proof. It may be OK to assume that is obvious. Similarly, it may be OK to state the "non-theorem :smile:" is also obvious for this example given that the function is constant on those two intervals. If you give the author the benefit of the doubt on what is obvious, then his example is OK.

    Normally in a problem like that, once you have covered the basics, you would assert continuity away from 0 since the formulas are polynomials, then show the left and right limits at zero are different, again using continuity of the left and right polynomials.

    I don't know if I really addressed your question. But I do think you understand what is going on and wouldn't be too concerned about it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook