First-order necessary condition for relative minimum point I'm studying linear and nonlinear programming and I came across with the following proposition : 
given $\rm x\in\Omega$ we are motivated to say  that a vector $\mathbf d$ is a feasible direction at $\mathbf x$ if there is an $\overline\alpha\gt0$ such that $\rm x+\alpha\mathbf d\in\Omega$ for all $\alpha$, $0\leqslant\alpha\leqslant\overline\alpha$.

I'm interested about the area I have highlighted with red rectangle. The result that $\nabla f(\textbf{x}^*) = \textbf{0}$ is intuitive, but how would one analytically show that indeed $\nabla f(\textbf{x}^*) = \textbf{0}$
 A: If $x^*$ is an interior point of $\Omega$, then every direction $d \in \mathbb{R}^n$ is a feasible direction from $x^*$, i.e., for any fixed $d$, $x^* + \alpha d \in \Omega$ for any sufficiently small values of $|\alpha|$. Defining $g(\alpha)$ as was done above, we again Taylor expand $g(\alpha)$ 
\begin{equation}
g(\alpha) = g(0) + g'(0)\alpha + \omicron(\alpha)
\end{equation}
Now let us assume that $g'(0) \neq 0$. Then, since we know that
\begin{equation}
\lim_{\alpha \to 0} \frac{\omicron(\alpha)}{\alpha} = 0,
\end{equation}
we know that given some $\epsilon > 0$, we can find an $\alpha$ (with $|\alpha|$ "small enough") such that
\begin{equation}
\left|\frac{\omicron(\alpha)}{\alpha}\right| < \epsilon.
\end{equation}
Let us choose
\begin{equation}
\epsilon = |g'(0)|.
\end{equation}
Then for small enough values of $|\alpha|$, we have
\begin{equation}
|\omicron(\alpha)| < \epsilon|\alpha| = |g'(0)\alpha|
\end{equation}
Going back to the first equation, we have
\begin{equation}
g(\alpha) < g(0) + g'(0)\alpha + |g'(0)\alpha|
\end{equation}
Since every direction is feasible, we can select $\alpha$ to have the opposite sign of $g'(0)$. Then $g'(0)\alpha = -|g'(0)\alpha|$ and the above equation becomes
\begin{equation}
g(\alpha) - g(0) < 0
\end{equation}
which contradicts $0$ being a minimum of $g$. Then $g'(0) = \nabla f(x^*)\cdot d = 0$. Since $d$ is arbitrary in this case, we conclude that $\nabla f(x^*) = 0$. 
The big difference between this proof and the proof above of the constrained case is that every direction is feasible here. This lets us always select $\alpha$ to have the opposite sign of $g'(0)$; whereas the proof you've provided considers $g'(0) < 0$ and small values of $\alpha > 0$, we consider both cases simultaneously here and rule out both $g'(0) > 0$ and $g'(0) < 0$ at the same time. 
