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Math Help - continuous

  1. #1
    Member vernal's Avatar
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    continuous

    Hi.

    let



    It is a continuous function f on x_0? is it true or false? please prove or example

    thanks
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  2. #2
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    Re: continuous

    Is there some reason why you don't want to do this problem? Just start looking at some examples.

    (I assume that f'_-(x_0) means \lim_{x\to x_0} f'(x) and equivalently for f'_+(x_0).)
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  3. #3
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    Re: continuous

    Quote Originally Posted by HallsofIvy View Post
    (I assume that f'_-(x_0) means \lim_{x\to x_0} f'(x) and equivalently for f'_+(x_0).)
    I would guess that f'_-(x_0)=\lim_{x\to0^-}\frac{f(x+h)-f(x)}{h} (left limit). Maybe the OP can clarify.
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  4. #4
    Member vernal's Avatar
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    Re: continuous



    if both of them there exists, then f countious or not?
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  5. #5
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    Re: continuous

    This could mean f'_+(x_0)=\lim_{x\to x_0^+}f'(x) or f'_+(x_0)=\lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}, which are two very different things. Regardless, though, you should make some effort in answering the question.
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  6. #6
    Member vernal's Avatar
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    Re: continuous

    Quote Originally Posted by emakarov View Post
    This could mean f'_+(x_0)=\lim_{x\to x_0^+}f'(x) or f'_+(x_0)=\lim_{x\to x_0^+}\frac{f(x)-f(x_0)}{x-x_0}, which are two very different things. Regardless, though, you should make some effort in answering the question.
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  7. #7
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    Re: continuous

    I knew it!

    It's a well-known fact that if f is differentiable at x_0, then f is continuous at x_0. Try to adapt the proof to this situation.
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  8. #8
    Member vernal's Avatar
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    Re: continuous

    Quote Originally Posted by emakarov View Post
    I knew it!

    It's a well-known fact that if f is differentiable at x_0, then f is continuous at x_0. Try to adapt the proof to this situation.
    thanks, but it is say f^' + and f^' - there exist, maybe f^' not exist.. Is it possible?
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  9. #9
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    Re: continuous

    Yes, for example for |x| at x = 0.

    Edit: But the proof of continuity is almost the same as in the case of the regular derivative.
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  10. #10
    Member vernal's Avatar
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    Re: continuous

    Quote Originally Posted by emakarov View Post
    Yes, for example for |x| at x = 0.

    Edit: But the proof of continuity is almost the same as in the case of the regular derivative.
    thanks for your help, and Excuse me. I'm not good English

    Well I did not answer my question
    Now if the f^' + and f^' - there exists, then f will be countions? is it true?

    i'm very sorry
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  11. #11
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    Re: continuous

    I realized that I myself got confused. No, if both one-sided derivatives exist, the function is not necessarily continuous. Both one-sided limits of the function exist (the proof of this is almost the same as for the fact that differentiable functions are continuous), but these limits are not necessarily the same. Consider an example shown here.
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