Show that if $x_n\rightarrow\ x_0^{-}$ and $y_n\rightarrow\ x_0^{+}$ then $ \lim \limits_{n \to \infty} \frac{f(y_n)-f(x_n)}{y_n-x_n} = f^\prime(x_0)$ 
Let $f: \Bbb[a,b]\rightarrow\Bbb{R}$ be differentiable in a point $a<x_0<b$. Use Taylor's expansion with the remainder of $f$ around $x_0$ to show that if $ x_n\rightarrow\ x_0^{-}$, and  $ y_n\rightarrow\ x_0^{+}$ then: 
  $$ \lim \limits_{n \to \infty} \frac{f(y_n)-f(x_n)}{y_n-x_n} = f^\prime(x_0)$$

I know that:
$f(x_n)= f(x_0)+f^\prime(x_0)(x_n-x_0) + R_1 (x_n) $
and:
$f(y_n)= f(x_0)+f^\prime(x_0)(y_n-x_0) + R_1 (y_n)$
Note: $R_1$ is the remainder. 
and then by substracting: 
$ \frac{f(y_n)-f(x_n)}{y_n-x_n} = f^\prime(x_0) + \frac{R_1(x_n)-R_1(y_n)}{x_n-y_n}$
but It seemed that I couldn't continue from here to anywhere, can someone please show the right way to do this? 

A nice exercise in Rudin's Principles of Mathematical Analysis (Exercise 5.19) states that, if $\alpha_n \to 0$ and $\beta_n \to 0$, then 
$$
D_n = \frac{f(\beta_n)-f(\alpha_n)}{\beta_n-\alpha_n}
$$
converges to $f'(0)$ under each of the following assumptions:


*

*$\alpha_n < 0 <\beta_n$;

*$0<\alpha_n<\beta_n$ and $\{\beta_n/(\beta_n-\alpha_n)\}_n$ is bounded;

*$f'$ is continuous.


Here you are in case 1. You can write
$$
D_n = \frac{f(\beta_n)-f(0)}{\beta_n-0}\frac{\beta_n}{\beta_n-\alpha_n} + \frac{f(\alpha_n)-f(0)}{\alpha_n}\frac{-\alpha_n}{\beta_n-\alpha_n}.
$$
Then you remark that $0<\frac{\beta_n}{\beta_n-\alpha_n}<1$ and $-1<\frac{\alpha_n}{\beta_n-\alpha_n}<0$. You should find your way from here.
By the way, case 3 is easily covered by some MVT or Taylor's theorem.
