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Let $f:I \to \mathbb{R}$ differentiable in the interval $I$ and let $a\in I$. For any sequences $(x_n)_{n\in \mathbb{N}}$ and $(y_n)_{n\in \mathbb{N}}$ in $I$ such that $\lim x_n = \lim y_n = a$, $x_n \neq y_n$, we have:

$$\lim \frac{f(y_n) - f(x_n)}{y_n - x_n} = f'(a) \tag{1}$$

Show that $f'$ is continuous at $a$.

The exercise belongs to the section of a book (Elon Lages Lima, Curso de Analise I) where it introduces the Lagrange mean value theorem, so it might be useful here, but I can't figure it out yet.

Some thoughts

This condition is stronger than differentiability. For example, the function: $$g(x)=\begin{cases}x^2\sin\left(\frac1x\right),&x\ne 0\\0,&x=0\end{cases}$$ is differentiable everywhere, but $g'$ is not continuous at $0$, hence not all pair of sequences $x_n \to 0, y_n \to 0, x_n \neq y_n$ will satisfy (1). Take for example $x_n=\frac1{(2n-\frac12)\pi}$ and $y_n=\frac1{(2n+\frac12)\pi}$, then the limit (1) converges to $-\frac{2}{\pi}$.

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closed as off-topic by colormegone, Claude Leibovici, user91500, user230715, JonMark Perry Jul 12 '16 at 9:32

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  • $\begingroup$ Something's not right here. What are you limits approaching? $\endgroup$ – KingDuken Jul 12 '16 at 2:14
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    $\begingroup$ Fix $\epsilon>0$. Can you write what you want to show, and then show some work towards that goal? In general, if you start with a sequence $x_n$ that approaches $a$, you will have the luxury of choosing $y_n$ as desired (perhaps to be "sufficiently close" to $x_n$ so some good things happen), just make sure $y_n$ also approaches $a$. By the way, I think you mean "$f'$ is continuous [at] $a$" rather than "$f'$ is continuous [in] $a$." $\endgroup$ – Michael Jul 12 '16 at 2:49
  • $\begingroup$ @Michael Fixed it. Thanks. $\endgroup$ – Michael Jul 12 '16 at 10:33
  • $\begingroup$ @Michael : Good observation about $x^2 \sin(1/x)$. To solve the problem, once you fix a sequence $x_n$ that approaches $a$, you want to show either that $\lim_{n\rightarrow\infty} |f'(a)-f'(x_n)| =0$, or (for a fixed $\epsilon>0$) that $|f'(a)-f'(x_n)|\leq \epsilon$ for all $x_n$ sufficiently close to $a$. Can you work towards that goal, using approximations for $f'(a)$ and $f'(x_n)$? $\endgroup$ – Michael Jul 12 '16 at 18:32
  • $\begingroup$ I got a solution on AoPS artofproblemsolving.com/community/u277018h1271060p6643173 since my post here was closed. Have a look if interested @Michael $\endgroup$ – Michael Jul 19 '16 at 23:52