If $f$ is differentiable at $a$, does it imply that $\lim\limits_{x,y\to a\atop x\ne y} \frac{f(x) - f(y)}{x-y} =f^{\prime}(a)$?

Question

Let $a \in \mathbb{R}$, $I \subseteq \mathbb{R}$ be a neighborhood of $a$, $f: I \rightarrow \mathbb{R}$ a function which is differentiable at $a$.

Want (either/or) :

• A function $f$ for which there exist sequences $(x_n), (y_n) \subset I$ such that $x_n \neq y_n \text{ }\forall n$ and $\lim\limits_{n \to \infty} x_n = a = \lim\limits_{n \to \infty} y_n$, but $\lim\limits_{n \to \infty} \frac{f(x_n) - f(y_n)}{x_n - y_n} = \xi \neq f^{\prime}(a)$ ;
• A function for which the quotient sequence simply diverges.
• An indication as to why what I'm asking for might not exist.

What I have managed to show:

Denoting $Q_{n} \equiv \frac{f(x_n) - f(y_n)}{x_n - y_n}$

1. Case 1: $$x_n < a < y_n \text{ } \forall n \implies Q_n \longrightarrow f^{\prime}(a)$$
2. Case 2: $$ \left( \frac{x_n - a}{x_n - y_n} \right), \left( \frac{y_n - a}{x_n - y_n} \right) \text{ bounded} \implies Q_n \longrightarrow f^{\prime}(a)$$

Answer 1

No. See for instance $a=0$, $I=\Bbb R$, $f(x)=\begin{cases}x^2\sin\frac1x&\text{if }x\ne 0\\ 0&\text{if }x=0\end{cases}$, $x_n=\frac1{2\pi n}$, $y_n=\frac1{2\pi n+e^{-n}}$.

Now, $f$ is differentiable with $$f'(x)=\begin{cases}0&\text{if }x=0\\ 2x\sin\frac1x-\cos\frac1x&\text{if }x\ne0\end{cases}.$$ By Lagrange's theorem, for all $n$ there is some $\xi_n\in(y_n,x_n)$ such that $\frac{f(x_n)-f(y_n)}{x_n-y_n}=f'(\xi_n)$. Therefore \begin{align}\limsup_{n\to\infty}\frac{f(x_n)-f(y_n)}{x_n-y_n}&\le\limsup_{n\to\infty}\sup_{y\in(y_n,x_n)} 2y\sin\frac1y-\cos\frac1y=\\& =\limsup_{n\to\infty}\sup_{z\in (2n\pi,2n\pi+e^{-n})} \frac2z\sin z-\cos z\le\\&\le \limsup_{n\to\infty}\frac2{2\pi n}-\cos(2\pi n+e^{-n})=\\&=\limsup_{n\to\infty}\frac1{\pi n}-\cos(e^{-n})=-1\\ \liminf_{n\to\infty}\frac{f(x_n)-f(y_n)}{x_n-y_n}&\ge\liminf_{x\to 0}f'(x)=-1\end{align}

Therefore $\lim_{n\to\infty}\frac{f(x_n)-f(y_n)}{x_n-y_n}=-1\ne f'(0)$.