I am currently reading "Variational analysis" by Rockafellar and Wets, and I am trying to understand the proof of Theorem 6.14 about normal cones to sets with constraint structure. But, I can't verify its discussion within the following image. In particular, I have no idea to show $$\langle y,\nabla F(\bar{x})(x-\bar{x})\rangle \leq o(|x-\bar{x}|)\tag{1}$$ from $$\langle y, F(x)-F(\bar{x})\rangle \leq o(|F(x)-F(\bar{x})|)\tag{2},$$ $$F(x)-F(\bar{x})=\nabla F(\bar{x})(x-\bar{x})+o(|x-\bar{x}|)\tag{3}.$$ By substituting (3) into (2) and dividing by $|x-\bar{x}|$, the right-hand side of (2) can be evaluated as $$\frac{o(|F(x)-F(\bar{x})|)}{|x-\bar{x}|}=\frac{o(|\nabla F(\bar{x})(x-\bar{x})+o(|x-\bar{x}|)|)}{|x-\bar{x}|}\leq \frac{o(|\nabla F(\bar{x})(x-\bar{x})|)}{|x-\bar{x}|}+\frac{o(|x-\bar{x}|)}{|x-\bar{x}|}\to \limsup_{x\to\bar{x}}\frac{o(|\nabla F(\bar{x})(x-\bar{x})|)}{|x-\bar{x}|}\ ({\rm as}\ |x-\bar{x}|\to 0).$$ However, I guess that $\frac{o(|\nabla F(\bar{x})(x-\bar{x})|)}{|x-\bar{x}|}$ doesn't go to $0$ except for $X \subset \mathrm{Ker}\nabla F(\bar{x})$. How do I show (1)?
Edit: Even if $F$ is (locally) Lipschitz continuous at $\bar{x}$ with $L>0$, we have $\frac{o(|F(x)-F(\bar{x})|)}{|x-\bar{x}|} \leq L$. Theorem 9.7 implies that a continuously differentiable mapping is strictly continuous, i.e., locally Lipschitz continuous, and thus $o(|F(x)-F(\bar{x})|)\neq o(|x-\bar{x}|)$. Is this discussion true?
ReEdit:
I was wrong. Since $F$ is locally Lipschitz, $o(|F(x)-F(\bar{x}|)=o(|x-\bar{x}|)$. Thank you.
I summarize my answer in Edit and ReEdit below.
The smoothness of $F$ yields $|F(x)-F(\bar{x})| = |\nabla F(\bar{x})(x-\bar{x})+o(|x-\bar{x}|)| \leq |\nabla F(\bar{x})|_{\mathrm{op}}|x-\bar{x}| + o(|x-\bar{x}|)$. Thus, $o(|F(x)-F(\bar{x})|)=o(|\nabla F(\bar{x})|_{\mathrm{op}}|x-\bar{x}| + o(|x-\bar{x}|)) + o(|\nabla F(\bar{x})|_{\mathrm{op}}|x-\bar{x}|)+o(o(|x-\bar{x}|))=o(|x-\bar{x}|)+o(|x-\bar{x}|)=o(|x-\bar{x}|)$.