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Analysis: Continuous Functions

  1. Oct 18, 2011 #1
    I did the work. I'm not sure on some of these.

    I think for (c) I need to make D = (0, infinity)

    http://i111.photobucket.com/albums/n149/camarolt4z28/1-3.png [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/2-3.png [Broken]

    http://i111.photobucket.com/albums/n149/camarolt4z28/3-1.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 18, 2011 #2

    micromass

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    Some remarks. You're defining sequences wrong. You're saying (for example):

    [tex]x_n=x[/tex]

    You can't do this. Your sequence must be dependent on n and not on x. So you need to write

    [tex]x_n=n[/tex]

    or

    [tex]x_n=\frac{1}{n}[/tex]

    Another remark. In (h), the function 1/f isn't well-defined. Indeed, it can happen that we divide by 0!!!! So you can't use that argument there (or at least not directly). Does it change anything if f(x)=0 for an x??

    For (c). You can't really do that. Indeed, [itex]f(+\infty)[/itex] isn't well-defined. You need c in the reals!!
     
  4. Oct 18, 2011 #3
    Yeah, I actually thought about that earlier. I knew I was being sloppy in my notation. Let me change that.

    Oh, in (c), I made an error. It should be pi/4. Is my approach correct?

    In (h), I said f(c) does not equal 0. I'm not really sure how to do this problem.
     
    Last edited: Oct 18, 2011
  5. Oct 18, 2011 #4

    micromass

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    The idea might be correct. But the use of infinite must be evaded.

    That's the point, you can't assume that f(c) doesn't equal zero!!! For all we know, f can be the zero function!! (hint hint)
     
  6. Oct 18, 2011 #5
    Why must the use of infinite be evaded? One of the definitions in a theorem for continuity says,

    If (xn) is any sequence in D such that (xn) converges to c, then limit of f(xn) as n tends to infinity is f(c). Isn't my example a valid counterexample?
     
  7. Oct 18, 2011 #6

    micromass

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    Your theorem applies if c is a real number. You can't choose c=infinity. Furthermore, f(infinity) is not well-defined.
     
  8. Oct 18, 2011 #7
    You're talking about (c)? What about (f)?

    For (c), it's not f(infinity). The argument is not infinity. Are you talking about (f)?
     
  9. Oct 18, 2011 #8

    micromass

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    In (c), what is your c?? What is f(c)??
     
  10. Oct 18, 2011 #9
    Haha. In (c), c = pi/4. f(c) is root2/2.

    I picked an f(c) value with a sequence that did not converge to that c. That's what the theorem states. If you know that f is continuous, then if a sequence converges to some c value, then the limit of f(of that sequence) will equal f(c).

    The statement is not necessarily true because it starts with the function value. It's backwards with respect to the theorem.
     
  11. Oct 18, 2011 #10

    micromass

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    Uuuh, that's not what I understand when I read (c). You better rephrase that answer...
     
  12. Oct 18, 2011 #11
    Really? Is what I just wrote correct? I thought I construed it properly mathematically on the paper.
     
  13. Oct 18, 2011 #12
    For (h), I suppose it's important to define a proper domain. If f is continuous at some point c and f(c) = 0 , then g(x) doesn't necessarily have to be continuous. But that's just a wild guess.
     
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