= Solution
For positive $\Delta_o\gg B$, expand the common radical as
$$
\sqrt{\Delta_o^2+18B\Delta_o+225B^2}
=\Delta_o+9B+\frac{72B^2}{\Delta_o}+O\!\left(\frac{B^3}{\Delta_o^2}\right).
$$
Substitution into the stated <crystal field theory> energies gives
$$
\boxed{\begin{aligned}
E_0&=-\frac45\Delta_o+3B-\frac{36B^2}{\Delta_o}+\cdots,\\
E_1&=\frac15\Delta_o,\\
E_2&=\frac15\Delta_o+12B+\frac{36B^2}{\Delta_o}+\cdots,\\
E_3&=\frac65\Delta_o.
\end{aligned}}
$$
The leading straight-line asymptotes retain the terms linear in $\Delta_o$ and constant in $B$. The $B^2/\Delta_o$ terms show from which side the interacting $T_{1g}$ levels approach them.
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