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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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    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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For every interval $J$, $f(J)=\{f(x)\mid x\in J\}$. Assume that $f$ is differentiable at $x_0$ with $f'(x_0)=c$ and assume without loss of generality that $c\geqslant0$. For every $\varepsilon\gt0$, there exists $\alpha\lt x_0\lt \beta$ such that for every $\alpha\lt x\lt\beta$, $ f(x_0)+c(x-x_0)-\varepsilon|x-x_0|\leqslant f(x)\leqslant f(x_0)+c(x-x_0)+\varepsilon|x-x_0|. $ In particular, for every $J=(a,b)\subseteq(\alpha,\beta)$, $(c_\varepsilon,d_\varepsilon)\subseteq f(J)\subseteq(a_\varepsilon,b_\varepsilon)$, with $ a_\varepsilon=f(x_0)+c(a-x_0)-\varepsilon(x_0-a),\qquad b_\varepsilon=f(x_0)+c(b-x_0)+\varepsilon(b-x_0), $ and $ c_\varepsilon=f(x_0)+c(a-x_0)+\varepsilon(x_0-a),\qquad d_\varepsilon=f(x_0)+c(b-x_0)-\varepsilon(b-x_0). $ Thus $d_\varepsilon-c_\varepsilon\leqslant|f(J)|\leqslant b_\varepsilon-a_\varepsilon$, that is, $(c-\varepsilon)(b-a)\leqslant|f(J)|\leqslant(c+\varepsilon)(b-a)$. Since $b-a=|J|$, this shows that $ \left|\frac{|f(J)|}{|J|}-c\right|\leqslant\varepsilon, $ for every interval $J\subseteq(\alpha,\beta)$. In this sense, $ \lim_{J\to x_0}\frac{|f(J)|}{|J|}=c. $

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This seems to be exercise 7.2.26 from the book Elementary Real Analysis by Brian S. Thomson,Judith B. Bruckner,Andrew M. Bruckner; p.278.

This exercise is contained in Section 7.2.3 The Derivative as a Magnification, where authors provide the following motivation for the derivative at a point.

If $J$ is a sufficiently small interval having $x_0$ as an endpoint, then the ratio $|f(J)|/|J|$ is approximately $|f'(x_0)|$, the approximation becoming "exact in the limit." Thus $|f'(x_0)|$ can be viewed as a "magnification factor" of small intervals containing the point $x_0$.


With this motivation we can formalize the above statement as: For very $\varepsilon>0$ and there exists $\delta>0$ such that for any interval $J$, such that $|J|<\delta$ and $J$ has $x_0$ as the endpoint, the inequality $\left|\frac{|f(J)|}{|J|}-|f'(x_0)|\right|<\varepsilon$ holds.

By the definition of derivative you have $\delta>0$ such that $|x-x_0|\le\delta$ implies $|\frac{f(x)-f(x_0)}{x-x_0}-f'(x_0)|<\varepsilon$. But this also implies $\left|\frac{|f(x)-f(x_0)|}{|x-x_0|}-|f'(x_0)|\right|<\varepsilon.$

If $J$ is interval of the form $[x_0,x]$ with $x_0-x<\delta$ and $y\in J$ then $\left|\frac{|f(y)-f(x_0)|}{y-x_0}-|f'(x_0)|\right|<\varepsilon\\ \left||f(y)-f(x_0)|-|f'(x_0)|(y-x_0)\right|<\varepsilon(y-x_0)\\ |f'(x_0)|(y-x_0)-\varepsilon(y-x_0) < |f(y)-f(x_0)| < |f'(x_0)|(y-x_0)+\varepsilon(y-x_0)\\ f(x_0)-|f'(x_0)|(y-x_0)-\varepsilon(y-x_0) < f(y) < f(x_0)+|f'(x_0)|(y-x_0)+\varepsilon(y-x_0) $ Considering the rightmost point of the interval (which fulfills $x-x_0=|J|$) we see that $|f(J)| \ge |f(x)-f(x_0)| \ge |f'(x_0)||J|-\varepsilon|J|.$ On the other hand, for any $y,y'\in J$ we have $f(y)-f(y')< |f'(x_0)|(y-y')+\varepsilon(y-x_0)+\varepsilon(y'-x_0) \le |f'(x_0)||J|+2\varepsilon|J|$. This implies $|f(J)|\le |f'(x_0)||J|+2\varepsilon|J|.$ Together we have $\left|\frac{|f(J)|}{|J|}-|f'(x_0)|\right| < 2\varepsilon$ for any interval $J$ which has endpoint $x_0$ and small enough length.