Show $ \lim_{n \to \infty} \frac{\ f(y_n)-f(x_n)}{ y_n-x_n} = f'(x_0)$ using Taylor Let $f:[a, b]\to R$ differentiable at $a<x_0<b$. Using taylor series show that if $x_n \to x_0^-$ and $y_n \to x_0^+$ then 
$$ \lim_{n \to \infty} \frac{\ f(y_n)-f(x_n)}{ y_n-x_n} = f'(x_0)$$
After I wrote the taylor expansion of 1's order I just returned to the definition of the derivative so that didn't help much...
Any other suggesions? Thanks
 A: For $n\in\mathbb N$ we have
$$ 
\begin{align}
\left|\frac{f(y_n) - f(x_n)}{y_n - x_n} - f'(x_0) \right|
&= \left| \frac{f(y_n) - f(x_0) - f'(x_0)(y_n - x_0) - (f(x_n) - f(x_0) - f'(x_0)(x_n - x_0))}{y_n - x_n} \right| \\
&\le \left| \frac{f(y_n) - f(x_0) - f'(x_0)(y_n - x_0)}{y_n - x_0} \right| + \left| \frac{(f(x_n) - f(x_0) - f'(x_0)(x_n - x_0)) }{x_n - x_0} \right| \\
&\to 0 + 0.
\end{align}
$$
A: I would like to mention that we are not using Taylor series to solve this problem. We are using Taylor expansions, and in particular we are using the first order Taylor expansion. A differentiable function is not guaranteed to have a Taylor series with positive radius of convergence, but it can be approximated by Taylor polynomials.
The first order Taylor expansion is given by $$f(x) = f(x_0) + f'(x_0)(x-x_0) + R(x)$$ where $\frac{R(x)}{x-x_0} \to 0$ as $x \to x_0$.
Now we examine $$\frac{f(y_n) - f(x_n)}{y_n - x_n} = \frac{f'(x_0)(y_n-x_0)-f'(x_0)(x_n-x_0) + R(y_n) - R(x_n)}{y_n-x_n}$$ $$=f'(x_0) + \frac{R(y_n) - R(x_n)}{y_n-x_n}$$
Thus in order to demonstrate the claim, we must show that $$\lim_{n\to\infty}\frac{R(y_n) - R(x_n)}{y_n-x_n} =0.$$
We need to convert this into an expression resembling $R(x)/(x-x_0)$, to claim that it goes to zero. This is where $y_n > x_0$ and $x_n < x_0$ comes in.
Notice that $$\left| \frac{R(y_n) -R(x_n)}{y_n-x_n} \right| \le \left|\frac{R(y_n)}{y_n-x_n} \right| + \left| \frac{R(x_n)}{y_n - x_n} \right|$$ and replacing $y_n - x_n$ with $y_n-x_0$ and $x_n - x_0$ only increases the fractions thus:
$$\left| \frac{R(y_n) -R(x_n)}{y_n-x_n} \right| \le \left| \frac{R(y_n)}{y_n-x_0} \right| + \left| \frac{R(x_n)}{x_n-x_0} \right| \to 0$$ as $n\to \infty$.
This last step is the crux of the matter, and it also appears in the answer provided by @user251257.
