This is a paragraph from Rudin PMA's solution manual, Exer 5.19.
Let $f$ be any function from $(-1,1)$ to $\Bbb R$ such that it is differentiable on $(-1,1)$ and, $\lim\limits_{x\to0}f'(x)$ does not exist. We know that $f'(x)$ does not tend to infinity as $x\to0$, since if it did, we would have $|f'(x)|>1+|f'(0)|$ for all sufficiently small nonzero $x$, and this contradicts the intermediate value property of derivatives. Hence there is a sequence $x_n\to0$, $x_n\ne0,$ such that $\lim\limits_{n\to\infty}f'(x_n)=L\ne f'(0)$.
Let $x_n$ and $y_n$ be such that $$0<|y_n-x_n|<\frac12|x_n|$$ and $$\left|\dfrac{f(y_n)-f(x_n)}{y_n-x_n}-f'(x_n)\right|<\dfrac{|L-f'(0)|}{2n}.\tag1$$ It is immediate that $$\lim\limits_{n\to\infty}\dfrac{f(y_n)-f(x_n)}{y_n-x_n}=L\ne f'(0).$$
Well, I cannot get how $\lim\limits_{n\to\infty}\dfrac{f(y_n)-f(x_n)}{y_n-x_n}=L\ne f'(0)$ follows.
Also, can we use an easier expression in $(1)$ on left side, for example $\dfrac{1}{n}$?
Given $\{x_n\}$ satisfying $\lim\limits_{n \to \infty} f'(x_n) = L \neq f'(0)$ and any $\{c_n\}$ such that $\lim\limits_{n \to \infty} c_n = 0$, if $\{y_n\}$ satisfies$$ \left| \frac{f(y_n) - f(x_n)}{y_n - x_n} - f'(x_n) \right| \leqslant |c_n|, \quad \forall n \in \mathbb{N}_+ $$ then$$ -|c_n| \leqslant \frac{f(y_n) - f(x_n)}{y_n - x_n} - f'(x_n) \leqslant |c_n|\\ \Longrightarrow f'(x_n) - |c_n| \leqslant \frac{f(y_n) - f(x_n)}{y_n - x_n} \leqslant f'(x_n) + |c_n|, $$ which implies$$ \varlimsup_{n \to \infty} \frac{f(y_n) - f(x_n)}{y_n - x_n} \leqslant \lim_{n \to \infty} f'(x_n) + \lim_{n \to \infty} |c_n| = L,\\ \varliminf_{n \to \infty} \frac{f(y_n) - f(x_n)}{y_n - x_n} \geqslant \lim_{n \to \infty} f'(x_n) - \lim_{n \to \infty} |c_n| = L. $$ Thus$$ \lim_{n \to \infty} \frac{f(y_n) - f(x_n)}{y_n - x_n} = L. $$ In particular, taking $c_n = \dfrac{L - f'(0)}{2n}$ to get the original proposition.