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? 

You need to use an estimate for the remainder. For exemple
$$R_1(x) = (x-x_0)\epsilon(x)$$
with $$\lim_{x\to x_0}\epsilon(x) = 0$$
Or in other words, $R_1(x) = o(x-x_0)$
