Question about a differentiable function at point $a$. Let $f$ be differentiable at point $a$. Prove than if $\lim \limits_{n \to \infty}x_n =\ a^{+}$ and $\lim \limits_{n \to \infty}y_n = a^{-}$ then $$\lim \limits_{n \to \infty} \frac{ f(x_n) - f(y_n)}{x_n - y_n} = f'(a)$$ 
I thought about proving this by using lagrange's theorem, but I didn't know where to go with this because the question has limit and its talking about one point, not an interval. 
Is there any ideas how to solve this question? 
 A: Since $f$ is differentiable on $a$ then 
$$f(x_n)=f(a)+(x_n-a)f'(a)+(x_n-a)\epsilon_1(x_n)$$
where $\epsilon_1(x_n)\xrightarrow{n\to\infty}0$ and similarly we have
$$f(y_n)=f(a)+(y_n-a)f'(a)+(y_n-a)\epsilon_2(y_n)$$
where $\epsilon_2(y_n)\xrightarrow{n\to\infty}0$. Now subtracting the two equalities and we get
$$f(x_n)-f(y_n)=(x_n-y_n)f'(a)+\underbrace{(x_n-a)\epsilon_1(x_n)-(y_n-a)\epsilon_2(x_n)}_{=R_n}$$
$$R_n=(x_n-y_n)\epsilon_1(x_n)+(y_n-a)(\epsilon_1(x_n)-\epsilon_2(y_n))$$
and notice that
$$0\le a-y_n=\underbrace{(a-x_n)}_{\le0}+(x_n-y_n)\le x_n-y_n$$ 
Can you take it from here?
A: Using algebraic manipulation we may write
$$\begin{align}\frac{f(x_n) - f(y_n)}{x_n - y_n} &= \frac{[f(x_n) - f(a)] - [f(y_n) - f(a)]}{x_n - y_n} = \frac{f(x_n) - f(a)}{x_n-y_n} - \frac{f(y_n) - f(a)}{x_n - y_n}\\&=\underbrace{\Bigg(\frac{x_n - a}{x_n - y_n}\Bigg)}_{t_n}\frac{f(x_n) - f(a)}{x_n-a} + \Bigg(\frac{(x_n - x_n) -(y_n - a)}{x_n - y_n}\Bigg)\frac{f(y_n) - f(a)}{y_n-a}\\&=t_n\underbrace{\frac{f(x_n) - f(a)}{x_n-a}}_{\to f'(a)} + \Big(1 - t_n\Big)\underbrace{\frac{f(y_n) - f(a)}{y_n-a}}_{\to f'(a)} \end{align}$$
Notice that $$y_n < a \Rightarrow x_n - a < x_n - y_n \Rightarrow \frac{x_n - a}{x_n - y_n} < 1$$ then the sequence $\{t_n\}$ is bounded and $0 < t_n < 1$. Then 
$$\lim_{n \to \infty}\frac{f(x_n) - f(y_n)}{x_n - y_n} = \lim_{n \to \infty}t_n\underbrace{\frac{f(x_n) - f(a)}{x_n-a}}_{\to f'(a)} + \Big(1 - t_n\Big)\underbrace{\frac{f(y_n) - f(a)}{y_n-a}}_{\to f'(a)} = f'(a)$$ 
