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Math Help - prove that f is differentiable at x_0 and that f' is continuous at x_0

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    prove that f is differentiable at x_0 and that f' is continuous at x_0

    suppose that f is continuous on (a,b). If f is also differentiable on (a,b), except possibly at a single point x_0 where lim_x-> x_0 f'(x) exists and is finite, prove that f is differentiable at x_0 and that f' is continuous at x_0.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    suppose that f is continuous on (a,b). If f is also differentiable on (a,b), except possibly at a single point x_0 where lim_x-> x_0 f'(x) exists and is finite, prove that f is differentiable at x_0 and that f' is continuous at x_0.
    f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}f'(x). Why?
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    Quote Originally Posted by Drexel28 View Post
    f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}f'(x). Why?
    not really. basically we are asked to prove \lim_{\delta\to 0}\lim_{h\to0}\frac{f(x_0+\delta+h)-f(x_0+\delta)}{h}=\lim_{h\to0} \lim_{\delta\to 0} \frac{f(x_0+\delta+h)-f(x_0+\delta)}{h}, and the conclusions will follow. to show that you can interchange the limits, probably some standard results should apply (it feels that uniform boundedness of f' is precisely what is needed)
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by choovuck View Post
    not really. basically we are asked to prove \lim_{\delta\to 0}\lim_{h\to0}\frac{f(x_0+\delta+h)-f(x_0+\delta)}{h}=\lim_{h\to0} \lim_{\delta\to 0} \frac{f(x_0+\delta+h)-f(x_0+\delta)}{h}, and the conclusions will follow. to show that you can interchange the limits, probably some standard results should apply (it feels that uniform boundedness of f' is precisely what is needed)
    Oh, is that so? What makes you think what I said was incorrect?
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    Quote Originally Posted by Drexel28 View Post
    f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}=\lim_{x\to x_0}f'(x). Why?
    b/c of the mean value theorem?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    b/c of the mean value theorem?
    Haha, well. Very indirectly. Think more "limity"
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    Quote Originally Posted by Drexel28 View Post
    Haha, well. Very indirectly. Think more "limity"
    hmm..I'm drawing a blank
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    hmm..I'm drawing a blank
    What is \lim_{x\to 0}\frac{\sin(x)}{x}? How do you know? I bet it isn't because you expanded it with a Maclaurin series or used a geometrically based squeeze theorem argument.
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    Quote Originally Posted by Drexel28 View Post
    What is \lim_{x\to 0}\frac{\sin(x)}{x}? How do you know? I bet it isn't because you expanded it with a Maclaurin series or used a geometrically based squeeze theorem argument.
    1? and im not sure if your using sarcasm lol but what about l'hopital's rule?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    about l'hopital's rule?
    Mhmm!
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    Quote Originally Posted by Drexel28 View Post
    Mhmm!
    hmm, i see that with the sinx/x but I'm confused at how to use it with the orig. problem lol
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    hmm, i see that with the sinx/x but I'm confused at how to use it with the orig. problem lol
    Look at L'hopital's rule. f(x)-f(x_0)\to 0 (by continuity), and x-x_0\to 0 (obviously) and so...
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    Quote Originally Posted by Drexel28 View Post
    Look at L'hopital's rule. f(x)-f(x_0)\to 0 (by continuity), and x-x_0\to 0 (obviously) and so...
    after rereading my question over and over again, I feel like just proving L'hopital's rule is what I have to do. but what about g?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    after rereading my question over and over again, I feel like just proving L'hopital's rule is what I have to do. but what about g?
    Try rereading it. The derivative exists everywhere except for maybe at that point, so L'hopital's applies and when you differentiate the numerator and denominator you get \lim_{x\to x_0}f'(x)...which...by...assumption...exists.
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    Quote Originally Posted by Drexel28 View Post
    Try rereading it. The derivative exists everywhere except for maybe at that point, so L'hopital's applies and when you differentiate the numerator and denominator you get \lim_{x\to x_0}f'(x)...which...by...assumption...exists.
    right, i understand that. But wait isn't it telling us that the limit as x->x_0 f'(x) does exist and is finite? Aye, im confusing myself, i think.
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