Proving that expression is equivalent to the definition of derivative Let $f$ be differentiable at $x=a$. 
Prove that if $x_n \to a^+$ and $y_n \to a^-$ then:
$$\lim_{n\to \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}=f'(a).$$
Every option that I think about seems to my very trivial, so I believe that I am doing something wrong. 
Both numerator and denominator approach zero as $n\to\infty$ as the case of the formal definition of derivative, but it isn't guaranteed that the limits are equal (“$\frac{0}{0}$”). 
Any direction?

First remark that the result is easy if you assume that $f$ is derivable with continuous derivative in a neighborhood of $a$. Indeed, by the mean value theorem, you can write $\frac{f(x_n)-f(y_n)}{x_n - y_n}$ as $f'(c_n)$ for some $c_n$ between $y_n$ and $x_n$ and the result follows by continuity of the derivative.
For the general case, I would do the following: write $\frac{f(x_n)-f(y_n)}{x_n - y_n}$ as $\frac{f(x_n) - f(a)}{x_n - a} \frac{x_n - a}{x_n - y_n} + \frac{f(a)-f(y_n)}{a-y_n} \frac{a-y_n}{x_n-y_n}$. Then we would like to compute the limits of these quantities and find that this gives $f'(a)$ by addition and multiplication.
The problem is the following: the quantity $\frac{x_n-a}{x_n - y_n}$ may well have no limit at all. For instance, take $a=0$, $x_n = 1/n$ and $y_n = -1/n$ for odd $n$ and $0$ for even $n$. Then this quantity is alternatively $1$ and $1/2$, hence does not converge.
Here is how I resolve this issue. Write $u_n$ for $\frac{f(x_n)-f(y_n)}{x_n - y_n}$. Then, $u_n$ is bounded since in the above writing, the quantites $\frac{x_n-a}{x_n - y_n}$ and $\frac{a-y_n}{x_n-y_n}$ are between $0$ and $1$. It is a standard fact that a bounded sequence converges if and only if it has a unique adherence value.
By Bolzano-Weierstrass theorem, let $u_{\phi(n)}$ be an extracted sequence that converges to $l_\phi$. By further extracting, we can also assume that the sequences $\frac{x_n-a}{x_n - y_n}$ and $\frac{a-y_n}{x_n-y_n}$ also converge to $l_x$ and $l_y$. Remark that necessarily, $l_x + l_y = 1$ since $\frac{x_n-a}{x_n - y_n} + \frac{a-y_n}{x_n-y_n} = 1$. By the above writing, this shows that $l_u = f'(a)l_x + f'(a)l_y = f'(a)$. Hence, the only adherence value of $u_n$ is $f'(a)$, showing that $f'(a)$ is the limit of $u_n$.
Maybe there is a shortcut to the extracting argument...
Edit: Indeed, there is a shortcut:
Write $u_n$ as $\frac{f(a)-f(y_n)}{a-y_n} + \frac{x_n - a}{x_n - y_n}(\frac{f(x_n) - f(a)}{x_n - a} - \frac{f(a)-f(y_n)}{a-y_n})$. The first term tends to $f'(a)$, both terms in the parenthesis have the same limit and the coefficient before the parenthesis is bounded.
