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Homework Help: Continuity function

  1. Sep 20, 2014 #1
    (a) State precisely the definition of: a function f is continuous at a point
    a ∈ R.

    (b) At which points x ∈R is the function:
    f(x) = sin(1/x)continuous?
    You may assume that g(x) = 1=x is continuous on its domain, and
    h(x) = sin(x) is continuous on its domain.

    (c) Let f and g be functions such that:
    1. g is continuous at 0, and g(0) = 0.
    2. For all x∈ R, lf(x)l <= lg(x)l.
    Use the
    ϵ− δ definition of limit to show that f is continuous at 0.

    I did part (a) and (b) but i don understand how to do part(c).Can any1 provide guidance on it?thanks
  2. jcsd
  3. Sep 20, 2014 #2
    First, write down precisely what the definition of continuity requires in order for an arbitrary function f to be continuous at 0. Then, if it is not already written in terms of ϵ− δ, translate the limit into the exact ϵ− δ definition of a limit. You should now have an ϵ− δ definition of what it would mean for an arbitrary function f to be continuous at 0. If you are unsure about any of this, write down exactly what you did for these parts here on the forum.
    Now, let f be any function that satisfies requirement (2) of part (c). Do you see anything in your definition that can now be simplified in this special case?
  4. Sep 20, 2014 #3
    What I did:
    For f to be continuous at 0, lim f(x) as x approaches to 0 =f(0).
    Thus, lim f(x) at x approaches to 0 exists and equal to f(0). If for every epsilon, there exists a delta such that if x satisfies
    0<lxl<delta then lf(x)-f(0)l<epsilon
    Can guide me on how to continue ?thank you
  5. Sep 20, 2014 #4
    Great. This is the statement that we would like to be true; that is, we want to prove that this statement is true for the particular function f described by precepts (1) and (2).
    So far, our definitions only involve one function: f. Precept (2) tells us that the particular function we want to consider must have the property that "##|f(x)| \leq |g(x)|## for all values of x". We would like to somehow replace all occurrences of f with g, since we know that g is continuous at 0 (precept (1)).
    Hmm. We know that we would like ##|f(x) - f(0)| < \epsilon##. We would like to claim that there is a value of ##\delta## that makes this inequality true. But wait. We do have a value of ##\delta## that makes g continuous at 0. Let us write out exactly what it means, in ##\epsilon-\delta## form, for g to be continuous at 0, given that g(0) = 0.
    Our plan is to use this inequality, along with the inequality relating f and g, to claim that there is a value of ##\delta## for which ##|f(x) - f(0)| < \epsilon##.
    Last edited: Sep 20, 2014
  6. Sep 20, 2014 #5
    Thanks for your guidance. Now, I reached lf(x)l<=lg(x)l<epsilon >>>>lf(x)l<epsilon... How should I continue?
  7. Sep 20, 2014 #6
    That's perfect. Now we know that for every value of ##\epsilon > 0##, there is some value of ##\delta > 0## such that, for all x, if ##x < \delta##, then ##|f(x)| < \epsilon##. We would just like to change that last phrase to ##|f(x) - f(0)| < \epsilon##.
    Remember that ##|f(x)| \leq |g(x)|## for all values of x. What does this imply about f(0) ?
  8. Sep 20, 2014 #7
    Can you tell me what is it or some hints?
  9. Sep 20, 2014 #8
    Since ##|f(x)| \leq |g(x)|## for all values of x, we must have ##|f(0)| \leq |g(0)|##. Since g(0) = 0, what does this imply about f(0) ?
  10. Sep 21, 2014 #9
    Thanks alot. It implies that f(0) must be 0 also...!thanks!
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