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Proof of $f'$ is continuous given some conditions

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Let $f:\mathbb{R}\to\mathbb{R}$ be a function, show the following two conditions are equivalent:

(1) $f$ is differentiable, with continuous derivative $f'$ on $\mathbb R$;

(2) For any $x_0\in\mathbb{R}$ and for any two sequences $\{x_n\},\{y_n\}$, convergent to $x_0$, such that $x_n \neq y_n$ for any positive integer $n$, the sequence $\left\{\frac{f(x_n)-f(y_n)}{x_n-y_n}\right\}$ is convergent.

For $(1)\implies (2)$: by Lagrange's MVT and continuous derivative, we have$$\frac{f(x_n)-f(y_n)}{x_n-y_n}=f'(\xi_n)\to f'(x_0).$$

For $(2)\implies (1)$: I can prove that $f'$ exists on $\mathbb R$ as follws:Take any $x_n>a$ and $x_n\to x_0$, $y_n=x_0$, then$$\lim_{n\to\infty}\frac{f(x_n)-f(x_0)}{x_n-x_0}$$exists, by Heine's Theorem,the right derivative $f_+'(x_0)$ exists. Similarly, the right derivative $f_-'(x_0)$ exists.For any $x_n<x_0,y_n>x_0$ and $x_n\to x_0,y_n\to x_0$, take$$z_n:\quad x_1,y_1,x_2,y_2,\cdots,x_n,y_n,\cdots,$$then$$\lim_{n\to\infty}\frac{f(z_n)-f(x_0)}{z_n-x_0}$$exists, and$$\frac{f(x_n)-f(x_0)}{x_n-x_0},\quad\frac{f(y_n)-f(x_0)}{y_n-x_0}$$ are subsequences of $\frac{f(z_n)-f(x_0)}{z_n-x_0}$, so $f_+'(x_0)=f_-'(x_0)$ and $f'(x_0)$ exists.Hence $f$ has derivative $f'$ on $\mathbb R$.

But, I'm stuck with the proof of continuity of $f'$. Can someone give some help or hints for the proof the continuity of $f'$. Any comments will welcome.


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