Let $F: \mathbb R^n \setminus \{ x^*\} \rightarrow \mathbb R$ be a continuous function defined everywhere except at a point $x^* \neq 0$. Assume that all directional limits of $F$ at $x^*$ exists, i.e,
$$ L(h) := \lim_{\epsilon \rightarrow 0} F(x^* + \epsilon h) \quad \text{exists in $\mathbb R$ for all} \quad h \neq 0. $$
Assume further that the finite minimum of $F$, denoted by $\lambda$, is only attained "at" $x^*$, i.e, $F(x) > \lambda, \forall x \in \mathbb R \setminus \{ x^*\}$ and there exists a sequence $x_n$ converging to $x^*$ such that $$ \lambda := \inf F = \lim_{n\rightarrow \infty} F(x_n).$$ Then is there an $h^*$ such that $\lambda = L(h^*)$?
An (incorrect) example of such function, with infimum also attained at (1,2), is $$ F(x,y) = \frac{3(x-1)^2 +(y-1)^2}{(x-1)^2 + (y-1)^2}, \quad (x^*,y^*) = (1,1), \quad \lambda = 1 = L((0,1)), \quad h^* = (0,1). $$
What I know is if $F$ is uniformly continuous then the answer is yes. However I think that when we remove uniformity then there should be a counter example. Thank you for your help.
Edit: Added "only attained at $x^*$" in the assumption on the infimum.