Let $\Omega = (a,b) \subseteq \mathbb{R}$, $f: \Omega \to \mathbb{R}$. We say that $f$ does NOT satisfy the "right-above right-below limit property" (my own naming for this) if and only if $\forall \delta > 0$ $\exists x_0,y_0 \in [x,x+\delta]$ such that $f(x_0) > f(x)$ ("right-upper") and $f(y_0) < f(x)$ ("right-lower"). In other words, it satisfies the property if "I can't take two different sequences from the right with one increasing to the value and one decreasing to the value."
Does $f$ continuous imply $f$ satisfies the "right-above right-below limit property?"
My guess is that this statement is false...looking at the Weierstrass function it would seem it does not satisfy this property...if so, could you explain where exactly it breaks down either with that or another counterexample (even arguing pictorially is fine)?
Finally, I came up with this characterization as a way of saying "given a continuous function and a point $x \in \Omega$, there exists a linear function $L:[x,x+\delta] \to \mathbb{R}$ such that $L \leq f|_{[x,x+\delta]}$." Is this "property" of mine equivalent to this "linear function dominance" notion?
Consider $$f(x) = \begin{cases} x \sin(1/x) & x \ne 0 \\ 0 & x = 0 \end{cases}$$ which is continuous and has $f(x_n) \downarrow 0$ for $x_n = \frac{1}{2 \pi n + \pi / 4}$ and $f(y_n) \uparrow 0$ for $y_n = \frac{1}{2 \pi n + 3 \pi / 4}$.