$a>0$. In other words for a given $\delta >0$ we have to show that $|\lambda(a)+M_{\cal A}|\le \delta$ as $a \rightarrow 0$ with $a>0$.

By the continuity of $\Psi_{p/q}$ we can choose $r_1>0$ so that if $|a|\le r_1$ and $|\e|\le r_1$ then $\bigl|\displaystyle\frac{1}{q}\Psi_{p/q}(x,a,\e)\bigr| \le \delta$ for any $x$. Choose $r_2>0$ using the continuity of $\gamma_{p/q}^-$ such that if $|a|<r_2$ then $|\gamma_{p/q}^-(a)-p/q|<r_1$. Now fix $r=\min(r_1,r_2)$ and take $0<a<r$.

For $\e(a)=a.\lambda(a)$, we are on $\gamma_{p/q}^-$ i.e. we are are on the tongue. Which implies that there is an $x_a$ such that $F_{p/q+\e(a),a}^{\circ q}(x_a) = x_a+p$. By Lemma \[lemma\_angle\] we see that $$\lambda(a)+\cal A_q(\p) (x_a)+\frac{1}{q}\Psi_{p/q}(x_a,a,\e(a))=0$$ $$\Rightarrow \lambda(a)=-\cal A_q(\p) (x_a)-\frac{1}{q}\Psi_{p/q}(x_a,a,\e(a)) \ge -M_{\cal A}-\delta$$ $$\Rightarrow \lambda(a)+M_{\cal A}\ge-\delta.$$

For the other inequality assume that $x_0$ is a point such that $\cal A_q(\p) (x_0)=M_{\cal A}$. Since we are considering the left boundary of the tongue $\cal T_{p/q}$, we note that the graph of the function $F_{p/q+\e(a),a}^{\circ q}$ lies below the graph of the function $y=x+p$. Thus $$F_{p/q+\e(a),a}^{\circ q}(x_0)\le x_0+p$$ $$\Rightarrow \lambda(a)+\cal A_q(\p) (x_0)+\frac{1}{q}\Psi_{p/q}(x_0,a,\e(a))\le0 \;\ (\mbox {By  Lemma \ref{lemma_angle}})$$ $$\Rightarrow \lambda(a)+M_{\cal A}\le -\frac{1}{q}\Psi_{p/q}(x_0,a,\e(a))\le \delta.$$

The other cases follow similarly. When the function $\cal A_q(\p)$ is constant, then $m_{\cal A} = M_{\cal A}=\cal M(\p)$. So, $$\gamma_{p/q}^{l}(a)=p/q-\cal M(\p)a+o(a)\quad \text{and}\quad 
\gamma_{p/q}^{r}(a)=p/q-\cal M(\p)a+o(a).$$

