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Math Help - differentiability and continuity

  1. #1
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    differentiability and continuity

    (a) Suppose that f: \Re\rightarrow\Reis differentiable and, that f(x)/x\rightarrow 0 as x\rightarrow\infty. Prove that, if f'(x)\rightarrow k as x\rightarrow\infty, then k=0.
    (b) Give an example of a function g:\Re\rightarrow\Re such that g(x)\rughtarrow 0 as x\rightarrow\infty, but g'(x) does not tend to a limit as x\rightarrow\infty



    (a) I know f(x) must increase slower that x. So for some N\in\Re, x_1,x_2\in\Re, |f(x_1)-f(x_2)|<|x_1-x_2|. Now the trouble of showing that this inequality increases until \frac{|f(x_1)-f(x_2)|}{x_1-x_2|}<\epsilon. Dunno what to do!
    (b) I tried all the functions I could think of. I figured it should be oscillating since we dont want the derivative to converge... But it doesnt seem to make since since a converged function/series isnt increasing or decreasing anymore. I thought of something like (-1)^x/x but the derivative is complex. Otherwise I was trying things like sin(1/x) but couldnt find anything that worked.
    Last edited by DontKnoMaff; November 18th 2010 at 10:37 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    a) it is not limitative to suppose that f(*) is differentiable in 0. In...

    http://www.mathhelpforum.com/math-he...tml#post582679

    ... it has been extablished that if f(*) is differentiable in 0 is...

     \displaystyle f(x)= f(0) + k\ x + x\ \varepsilon (x) , \lim_{x \rightarrow 0} \varepsilon (x) =0 (1)

    From (1) we derive...

    \displaystyle f^{'}(x)= k + \varepsilon (x) + x\ \varepsilon^{'} (x) (2)

    ... and...

    \displaystyle \frac{f(x)}{x}= k + \varepsilon (x) (3)

    ... so that if \displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x} = k for (3) is \displaystyle \lim_{x \rightarrow \infty} \varepsilon(x)=0 , if \displaystyle \lim_{x \rightarrow \infty} f^{'} (x)= k for (2) is \displaystyle \lim_{x \rightarrow \infty} x\ \varepsilon^{'} (x)=0 and [finally...] if \displaystyle \lim_{x \rightarrow \infty} \frac{f(x)}{x} = 0 that means that is k=0...

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 19th 2010 at 04:53 AM.
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  3. #3
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    any help on part b) anybody?
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  4. #4
    MHF Contributor chisigma's Avatar
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    If the Laplace Tranform \mathcal {L} \{\varphi(t)\}= f(s) exists , then the 'Final value theorem' extablishes that...

    \displaystyle \lim_{t \rightarrow \infty} \varphi(t) = \lim_{s \rightarrow 0} s\ f(s) (1)

    Now is...

    \displaystyle \mathcal{L} \{\varphi^{'} (t)\} = s\ f(s) - \varphi(0) (2)

    ... and taking into account (1) and the fact that is \displaystyle \lim_{t \rightarrow \infty} \varphi(t)=0 is...

    \displaystyle \lim_{t \rightarrow \infty} \varphi^{'}(t) =  \lim_{s \rightarrow 0} \{ s^{2}\ f(s) - s\ \varphi(0) \} =0 (3)

    Kind regards

    \chi \sigma
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  5. #5
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    Okay, f(x)=\frac{\sin(x^3)}{x} works.
    Last edited by Jose27; November 22nd 2010 at 07:48 PM.
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  6. #6
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by chisigma View Post
    If the Laplace Tranform \mathcal {L} \{\varphi(t)\}= f(s) exists , then the 'Final value theorem' extablishes that...

    \displaystyle \lim_{t \rightarrow \infty} \varphi(t) = \lim_{s \rightarrow 0} s\ f(s) (1)

    Now is...

    \displaystyle \mathcal{L} \{\varphi^{'} (t)\} = s\ f(s) - \varphi(0) (2)

    ... and taking into account (1) and the fact that is \displaystyle \lim_{t \rightarrow \infty} \varphi(t)=0 is...

    \displaystyle \lim_{t \rightarrow \infty} \varphi^{'}(t) =  \lim_{s \rightarrow 0} \{ s^{2}\ f(s) - s\ \varphi(0) \} =0 (3)

    Kind regards

    \chi \sigma
    In Gluskin’s paper ‘Let us teach this generalization of the final-value theorem’ [http://iopscience.iop.org/0143-0807/24/6/005/] there are some important details that usually You don’t find in ordinary complex analysis books. Among others in cap. 4 is written…

    If \displaystyle \lim_{t \rightarrow \infty} f(t) and the Laplace Transform F(s) = \mathcal{L} \{f(t)\} exist, then we have the ‘final value theorem…

    \displaystyle \lim_{t \rightarrow \infty} f(t) = \lim_{s \rightarrow 0} s\ F(s) (4)

    For this equality to be true we assume that f(*) has derivative bounded in (0,\infty) and absolutely integrable in [0,\infty).

    For these reasons what i wrote is valid only under the conditions reported by Gluskin...

    Kind regards

    \chi \sigma
    Last edited by chisigma; November 23rd 2010 at 12:16 PM. Reason: The post has been fully revised...
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