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Wallach

  • 2 years ago

I have the following function f(x)=sin(1/x) I've been told that the function is continuous for any x. I understand that the above function is continuous at (0,infinite) and also at (-infinite,0) but as much as I know the limit of this function while x -->0 does not exist or is undefined so how come the function is still continuous??

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  1. terenzreignz
    • 2 years ago
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    Perhaps it's for all x in its domain?

  2. experimentX
    • 2 years ago
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    the function is not continuous at x=0, are you missing something?

  3. Wallach
    • 2 years ago
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    it's for \[f(x) \rightarrow R:R\]

  4. experimentX
    • 2 years ago
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    if f(x) = x sin(1/x), the function could have been made continuous ... you might be interested in http://math.stackexchange.com/questions/199603/is-exp-left-1-over-x2-right-differentiable-at-x-0

  5. Wallach
    • 2 years ago
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    I might be missing some thing but I'm not sure what I would have uploaded the question but it's in Hebrew

  6. Wallach
    • 2 years ago
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    that I know. that's because x while x-->0 equals 0 and sin(1/x) is a Bounded function

  7. Wallach
    • 2 years ago
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    sry I think that the right way to translate the question is: does the Intermediate value theorem is true for sin(1/x) and if it does (and it does) explain

  8. experimentX
    • 2 years ago
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    bounded does not necessarily mean it is continuous. as x approaches to 0 the function oscillates violently between -1 and 1 without converging to any particular value. I don't think sin(1/x) is continuous at 0. Both because limit and value are not exactly defined.

  9. Wallach
    • 2 years ago
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    that's what I thought

  10. experimentX
    • 2 years ago
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    f(x) = x sin(1/x) is particularly interesting ... being continuous but not differentiable at x=0.

  11. Wallach
    • 2 years ago
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    but before I was talking about the following statement: when given the following limit \[\lim_{x \rightarrow 0} f(x)*g(x) = ?\] if \[\lim_{x \rightarrow 0} f(x) = 0\] AND g(x) is bounded then \[\lim_{x \rightarrow 0} f(x)*g(x) = 0\]

  12. Wallach
    • 2 years ago
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    which is the case that you're presenting

  13. experimentX
    • 2 years ago
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    yeah ... the case is different from what you are asking.

  14. Wallach
    • 2 years ago
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    but lets go back to this for a sec : does the Intermediate value theorem is true for sin(1/x) and if it does (and it does) explain because if think this is the correct translation to the question I was given

  15. experimentX
    • 2 years ago
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    the intermediate value theorem only applies for continuous function on some interval. Though for this particular Q i can say exactly that f(0) is between -1 and 1.

  16. Wallach
    • 2 years ago
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    right that's the thing the title of the question was is the Intermediate value theorem correct only for continuous function?? here are a few functions etc. etc.

  17. Wallach
    • 2 years ago
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    I think I understood the whole idea behind the question and I don't want to take too much of your time but any way thank you for your help :)

  18. experimentX
    • 2 years ago
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    let me ask this to experts!! I'll reply back later.

  19. Wallach
    • 2 years ago
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    tnk u

  20. experimentX
    • 2 years ago
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    http://chat.stackexchange.com/transcript/message/6829811#6829811

  21. Wallach
    • 2 years ago
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    yes I know there's no f(0) that's why the function is not continuous but why is it true that the Intermediate value theorem is still true for this function?

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