-1
$\begingroup$

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 $$\lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0} = f'(x_0).$$

  • 0
    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
  • 0
    using the definition of limits i think. where f(x) - f(x0)/x - x02012-11-01
  • 0
    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
  • 0
    Maybe you meant $|J|\to 0$ where $J$ is an interval containing $x_0$?2012-11-01
  • 0
    Please consider rewriting your post to mention what you tried and to cancel all those imperatives.2012-11-01
  • 0
    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
  • 0
    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

2 Answers 2

2

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, $$ \lim_{J\to x_0}\frac{|f(J)|}{|J|}=c. $$

0

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.