Let $f$ continuos in $a\in I$. If $\lim \frac{f(y_n)-f(x_n)}{y_n-x_n}=L$ then $f'(a)=L$ Let $f:I\rightarrow \mathbb{R}$ continuos in $a\in I$ interior point. If
$$\lim \frac{f(y_n)-f(x_n)}{y_n-x_n}=L$$
for every $(x_n)$, $ (y_n)$ with $x_n<a<y_n$ and $\lim x_n=\lim y_n =a$ then $f'(a)=L$.
I know that if $f'(a)$ exist then
$$f'(a)=\lim \left[\frac{f(a+\frac{1}{n})-f(a-\frac{1}{n})}{\frac{2}{n}}\right]=L.$$
But I don't know how to profe that $f'(a)$ exist using the continuity of $f$ in $a$.
 A: Let $(x_n)$ be any sequence with $x_n < a$ and $\lim x_n = a$. Now for each $n \in \Bbb{N}$, notice that
$$ y \mapsto \frac{f(x_n) - f(y)}{x_n - y} $$
is continuous at $y = a$. So we can pick $y_n \in I$ such that $a < y_n < a+n^{-1}$ and
$$ \left| \frac{f(x_n) - f(y_n)}{x_n - y_n} - \frac{f(x_n) - f(a)}{x_n - a} \right| < \frac{1}{n}. $$
Then it follows that
$$ \left| \frac{f(x_n) - f(a)}{x_n - a} - L \right| \leq \left| \frac{f(x_n) - f(y_n)}{x_n -y_n} - L \right| + \frac{1}{n}. $$
and thus we get
$$ \lim_{n\to\infty} \frac{f(x_n) - f(a)}{x_n - a} = L. $$ This shows that $f_{-}'(a) = L$. Similar argument shows that $f_{+}'(a) = L$ as well.
A: This can be proven by contradiction. Here's an outline of how this would go: Suppose that it is not the case that $f'(a)=L$. Then $\frac{f(x)-f(a)}{x-a}$ does not have a limit of $L$ as $x$ approaches $a$. Therefore, there is an $\epsilon>0$ such that for every $\delta>0$, there is some $x$ with $0<|x-a|<\delta$ such that $\left|\frac{f(x)-f(a)}{x-a}-L\right| \geq \epsilon$. Let $(\delta_n)$ be any sequence of positive real numbers converging to zero (e.g., take $\delta_n=1/n$). Then for each $n$, there is some $x_n$ with $0<|x_n-a|<\delta_n$ such that $\left|\frac{f(x_n)-f(a)}{x_n-a}-L\right| \geq \epsilon$. Now, either there are infinitely many $n$ such that $x_n<a$ or there are infinitely many $n$ such that $x_n>a$. Without loss of generality, assume there are infinitely many $n$ such that $x_n<a$. By restricting to the subsequence of $x_n$ where $x_n<a$ (and restricting to the corresponding subsequence of $\delta_n$), we may assume that $x_n<a$ for all $n$. Now, for fixed $n$, $\frac{f(x_n)-f(y)}{x_n-y}$ is a continuous function of $y$ at $y=a$, so we may choose $y_n$ with $a<y_n<a+\delta_n$ with $$\left|\frac{f(x_n)-f(y_n)}{x_n-y_n}-\frac{f(x_n)-f(a)}{x_n-a}\right|<\epsilon/2.$$
Then $x_n<a<y_n$ and $\lim x_n=\lim y_n=a$, and by the triangle inequality, 
$$\left|\frac{f(x_n)-f(y_n)}{x_n-y_n}-L\right|\geq \epsilon$$
for all $n$, contrary to the assumption.
