Let $G$ a connected graph. If $x$ is a cut vertex of $G$, then $\lambda(G)\le\dfrac{\delta(x)}{2}$. Let $G$ a connected graph. If $x$ is a cut vertex of $G$, then $\lambda(G)\le\dfrac{\delta(x)}{2}$.
Where $\lambda(G)=\min\{|A| : A \text{ is a cut set of edges}\}$
It is not clear to me how to demonstrate that result. I know that the cardinal of the set of edges whose one of its end points is x, $|E_G(x)|=\delta(x)$ and this set is a cut set of edges, then $$\lambda(G)\le\delta(x)$$
 A: Let's assume that we generate 2 connected components by removing the vertex $x$. denote these components as $G_1$ and $G_2$, define $\delta_i(x)$ as the number of edges between $x$ and $G_i$, $i = 1,2$. Clearly $\delta_1(x) + \delta_2(x) = \delta(x)$.
If $\delta_1(x) < \delta_2(x)$ then we can remove $\delta_1(x) < \delta(x) / 2$ edges from $G$ to disconnect it from $G_1$. Which means $\lambda(G) \leq \delta_1(x) < \delta(x) / 2$, where $\lambda(G) \leq \delta_1(x)$ because it can be another cut vertex with less edges. Note that we don't use $\delta_2(x)$ because we remove less edges by considering $\delta_1(x)$ and we are searching for a minimum.
In the worst case, $\delta_1(x) = \delta_2(x)$ which would mean $\lambda(G) \leq \delta_1(x) = \delta(x) / 2$.
By the above reasoning, we have proved the proposition for the case where $x$ generates 2 connected components. If $x$ generates $n$ components $G_1, ..., G_n$, then again if one component has more edges to $x$ than any of the other, we can choose the edges from the component with the least number of connections to $x$, which should be less than or equal to $\delta(x) / n \leq \delta(x) / 2$ (observe that $n =1$ is not a valid case).
Thus, as $x$ is an arbitrary cut vertex we can only ensure $\lambda(G) \leq \delta(x) /2$.
