Prove that $f$ is differentiable One exercise from a list. I have no idea how to finish it.

Let $I=[c,d]\subset \mathbb{R}$.
Let $f:I\to \mathbb{R}$ be continuous at $a\in (c,d)$.
Suppose that there exists $L\in \mathbb{R}$ such that $$\lim \frac{f(y_n)-f(x_n)}{y_n-x_n}=L$$ for every pair of sequences $(x_n),(y_n)$ in $I$, with $x_n<a<y_n$ and $\lim x_n=\lim y_n=a$.
Prove that $f'(a)$ exists and it is equal to $L$.

Any help? Thanks in advance.
 A: This proof uses the idea I indicated in my comment, even though it may not look like it.
I claim that
$$\frac{f(y_n) - f(a)}{y_n-a} \to L$$
for every sequence $(y_n)$ satisfying $y_n>a$ and $y_n\to a$. Indeed, given $y_n>a$ choose $x_n<a$ so that $|x_n-a|<1/n$ and
$$\left|\frac{f(y_n)-f(x_n)}{y_n-x_n} - \frac{f(y_n)-f(a)}{y_n-a}\right|<\frac{1}{n}.$$
(This uses the assumed continuity of $f$.) Now let $n\to\infty$ in this inequality.
Similarly we can prove that 
$$\frac{f(a) - f(x_n)}{a-x_n} \to L$$
for every sequence $(x_n)$ such that $x_n<a$ and $x_n\to a$.
A: Maybe this is one of these rare cases when reasoning by contradiction helps. Thus, assume without loss of generality that $L=0$ (otherwise, replace $f$ by $x\mapsto f(x)-Lx$) and that the conclusion is false.  Thus, either $f'(a)$ does not exist or $f'(a)\ne 0$, in any case, there exists a sequence $(z_n)_n$ and some $\varepsilon\gt0$ such that $z_n\ne a$, $z_n\to a$, and $|f(z_n)-f(a)|\geqslant2\varepsilon|z_n-a|$ for every $n$.
Assume without loss of generality that $z_n\lt a$ infinitely often and define $(x_n)_n$ as the subsequence of $(z_n)_n$ made of the terms $\lt a$ hence $x_n\lt a$, $x_n\to a$, and $|f(x_n)-f(a)|\geqslant2\varepsilon(a-x_n)$. 
Now, $f(x)\to f(a)$ when $x\to a$, $x\gt a$, hence, for each $n$, one can choose $y_n\gt a$ such that $|f(y_n)-f(a)|\leqslant\varepsilon(a-x_n)$ and $y_n-a\leqslant\varepsilon(a-x_n)$. In particular, $(1+\varepsilon)(a-x_n)\geqslant y_n-x_n$. One gets
$$
|f(y_n)-f(x_n)|\geqslant|f(x_n)-f(a)|-|f(y_n)-f(a)|\geqslant\varepsilon(a-x_n)\geqslant(\varepsilon/(1+\varepsilon))(y_n-x_n),
$$
for every $n$, and, in particular,
$$
\frac{f(x_n)-f(y_n)}{y_n-x_n}\not\to0.
$$
A: The idea is that since the symmetric limit converges for every pair of sequences, we pick two sequences which show left and right convergence. If we define one sequence to be much closer to $a$ than the other, it becomes almost equivalent to approaching through the latter. 
The below is more of a sketch than a proof.
Since $f$ is continuous, for each $n\in\mathbb{N}$ there exists $\delta_{n} > 0$ such that $$\vert x - a \vert < \delta_n \implies \vert f(x) - f(a) \vert < \frac{1}{n^2}$$
Let $a - \min\left(\delta_n,\frac{1}{n^2}\right)$ define $x_n$ and let $y_n = a + \frac{1}{n}$
$$L=\lim_{n\rightarrow\infty}\frac{f(y_n) - f(x_n)}{y_n - x_n}$$
$$=\lim_{n\rightarrow\infty}\frac{f(a + \frac{1}{n}) - f(a)}{\frac{1}{n} + \mathcal{O}\left(\frac{1}{n^2}\right)} + \frac{f(a) - f(x_n)}{\frac{1}{n} + \mathcal{O}\left(\frac{1}{n^2}\right)}$$
$$=\lim_{n\rightarrow\infty}\frac{f(a + \frac{1}{n}) - f(a)}{\frac{1}{n} + \mathcal{O}\left(\frac{1}{n^2}\right)} + \frac{\mathcal{O}\left(\frac{1}{n^2}\right)}{\frac{1}{n} + \mathcal{O}\left(\frac{1}{n^2}\right)}$$
$$=\lim_{n\rightarrow\infty}\frac{f(a + \frac{1}{n}) - f(a)}{\frac{1}{n} + \mathcal{O}\left(\frac{1}{n^2}\right)} = f'(a)$$
The argument should probably be made a bit more rigorous and general, in particular you must show that this convergence holds for every $y_n$ which will involve changing the delta argument a little bit (still requiring $x_n - a \ll y_n - a$ for whatever $y_n$ you should choose). But the overall idea should hold.
