and $g^{k}$ is $H^{k}$-free. Then $$E(G^{k})=\{F\in \binom{[n]}{k}:\text{either } F\in E(g^{k}) \text{ or } F\cap \bar{X}\neq \emptyset\}.$$ Then $G=G^{1}\cup G^{k}$ is $H$-free and (by choice of $x$) we have that $\displaystyle \lim_{n\to\infty}h_{n}(G)$ attains the upper bound of $\pi(H)$. $\square$

Let us now return to the task of determining $\pi(H)$ when $H=H^{1}\cup H^{2}$.

Let $H=H^{1}\cup H^{2}$. If $H^{2}$ is not bipartite, then $$\pi(H)=1+\pi(H^{2})= 1+\left(1-\frac{1}{\chi(H^{2})-1}\right)=2-\frac{1}{\chi(H^{2})-1}.$$

[**Proof:**]{} First, $\pi(H)\geq 1+\pi(H^{2})$ since one can construct an $H$-free graph $G_{n}$ by letting $$E(G_{n})=\{\{v\}:v\in V(G_{n})\}\cup E(G^{\prime}_{n})$$ where $G^{\prime}_{n}$ attains $h_{n}(G^{\prime}_{n})=\pi_{n}(H^{2})$ and $G^{\prime}_{n}$ is $H^{2}$-free. Then $$\pi(H)\geq \lim_{n\to\infty} h_{n}(G_{n}) = \lim_{n\to\infty} 1+\pi_{n}(H^{2}) = 1+\pi(H^{2}).$$

To get the upper-bound, first add every missing $1$-edge into $H$, call the new graph $H^{\prime}$. Note that $\pi(H)\leq \pi(H^{\prime})$. Note that we didn’t change the edge set $H^{2}$. The Erdős-Stone-Simonovits theorem states that if $H^{2}$ is not bipartite, then $\pi(H^{2})=1-\frac{1}{\chi(H^{2})-1}$. Also, if $H^{2}$ is not bipartite, then $\chi(H^{2})\geq 3$. With the added vertices, taking $k=2$, we apply the previous theorem. Since $$\pi(H^{2})=1-\frac{1}{\chi(H^{2})-1}\geq 1-\frac{1}{2}$$ we may conclude that $\pi(H)\leq \pi(H^{\prime})=1+\pi(H^{2})$. $\square$

It remains to investigate the cases when $H^{2}$ is bipartite.

Let $H=H^{1}\cup H^{2}$. If $H^{2}$ is bipartite and $K_{2}^{\{1,2\}}\subseteq H$ then $\pi(H)=\frac{5}{4}$.

[**Proof:**]{} First, in example 1, we computed $\pi(K_{2}^{\{1,2\}})=\frac{5}{4}$. Second, $H$ must 