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Let $f$ be differentiable on an interval $I$ and let $x_0$ be an interior point of $I$. Make precise the following statement and prove it: $$\lim_{J \to x_0} \frac{|f(J)|}{|J|} = |f '(x_0)|$$

using the definition of limits where $$\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = f'(x_0).$$

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    consider $f(x) = x^2$, and $x_0 = 1$. Then $$\lim_{J \rightarrow x_0} \frac{|x^2|}{|x|} = \lim_{J \rightarrow 1} |J| = 1 \neq 2 = |f'(1)|$$so maybe you want a different statement?2012-11-01
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    using the definition of limits i think. where f(x) - f(x0)/x - x02012-11-01
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    The first statement doesn't make sense as it stated. But one can understand what is meant to be here. For example $J\to x_0$ is means interval $J$ contracts to the point $x_0$, $|f(J)|$ and $|J|$ means lengths of respective intervals.2012-11-01
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    Maybe you meant $|J|\to 0$ where $J$ is an interval containing $x_0$?2012-11-01
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    Please consider rewriting your post to mention what you tried and to cancel all those imperatives.2012-11-01
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    This seems to be exercise 7.2.26 from the book [Elementary Real Analysis](http://classicalrealanalysis.info/) by Brian S. Thomson,Judith B. Bruckner,Andrew M. Bruckner; [p.278](http://books.google.com/books?id=vA9d57GxCKgC&pg=PA278).2012-11-01
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    The original version appears to be identical to [this Yahoo! Answers question](http://answers.yahoo.com/question/index?qid=20121031205358AAYPjs8), errors and all.2012-11-01

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