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

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: $$ \left( \frac{x_n - a}{x_n - y_n} \right), \left( \frac{y_n - a}{x_n - y_n} \right) \text{ bounded} \implies Q_n \longrightarrow f^{\prime}(a)$$