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Thread: Limit and Derivative proof

  1. #1
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    Limit and Derivative proof

    Let $\displaystyle f0,1) \rightarrow \mathbb {R} $ be differentiable and let $\displaystyle f'(x)$ be Bounded. Show that $\displaystyle \lim _{x \rightarrow 0^+ } f(x) $ exist.

    Idea so far.

    So if I define a sequence that converges to a point at 0, f would be able to able to map that into the co-domain, would that be the right start?

    Thanks.
    Last edited by tttcomrader; Oct 17th 2008 at 12:15 PM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let $\displaystyle f0,1) \rightarrow \mathbb {R} $ be differentiable and let $\displaystyle f'(x)$ be founded. [I'll assume that should be "bounded".] Show that $\displaystyle \lim _{x \rightarrow 0^+ } f(x) $ exist.

    Idea so far.

    So if I define a sequence that converges to a point at 0, f would be able to able to map that into the co-domain, would that be the right start?
    Yes, that is a good way to start. Now prove that the sequence $\displaystyle f(x_n)$ is Cauchy, and hence has a limit. Then you'll need to check that this limit is the same for all sequences $\displaystyle x_n\to0$.

    (A quicker way to say the same thing is that a function with a bounded derivative is uniformly continuous; and if a function is uniformly continuous on a set A then it extends continuously to the closure of A.)
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    So in a differentiable function f, with f'(x) bounded, any sequence would be cauchy if it is mapped by f?
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    Quote Originally Posted by tttcomrader View Post
    So in a differentiable function f, with f'(x) bounded, any sequence would be cauchy if it is mapped by f?
    Yes, by the mean value theorem. If $\displaystyle |f'(x)|\leqslant M$ for all x, then $\displaystyle f(y)-f(x)=f'(\xi)(y-x)$ for some ξ between x and y, and so $\displaystyle |f(y)-f(x)|\leqslant M|y-x|$. If $\displaystyle (x_n)$ is Cauchy then $\displaystyle |f(x_m)-f(x_n)|\leqslant M|x_m-x_n|$, from which it follows that $\displaystyle (f(x_n))$ is also Cauchy.
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    Okay, so I'm picking up from after I know that $\displaystyle f(x_n)$ converges.

    Now, suppose that $\displaystyle f(x_n) \rightarrow L $.

    Since f is differetiable on (0,1), it is also continuous on (0,1), implies that $\displaystyle f(x_n) \rightarrow f(0) $ since $\displaystyle x_n \rightarrow 0 $.

    So is this enough to show that $\displaystyle \lim _{x \rightarrow 0^+} f(x) $ exist and it is also equals to $\displaystyle f(0)$?
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    Please look at Prof. Opalg’s second comment about uniform continuity.
    A function with a bounded derivative is uniformly continuous.
    Now go back to the last question I helped you with.
    I think that was a lemma for this question.
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    let $\displaystyle \epsilon > 0 $ be given, and find $\displaystyle \delta > 0 $, and suppose that the sequence $\displaystyle \{ x_n \}$ converges to 0. Then $\displaystyle \exists N \in \mathbb {N} $ such that whenever $\displaystyle n \geq N $ implies that $\displaystyle d(x_n,0)< \delta $

    Since f is uniformly continuous, we have $\displaystyle p(f(x_n),f(0)) < \epsilon $

    So for all sequences that converges to 0, the sequences after being mapped by the function f would converge to f(0).

    I hope this is right, suicide or not.

    Thanks!!!
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