Solution
= Solution
Using the <Taylor expansion>
$$
\log(1+ae^{-a})=a-\frac32a^2+O(a^3),
$$
we obtain
$$
t^*=3-\frac32a+O(a^2),
\qquad
\boxed{t^*\longrightarrow3\quad\text{as }a\to0}.
$$
When cooling is negligible, the milk lowers the temperature by one unit at $t=1$. The hot-water term begins at $t=2$ and raises it at one unit per unit time, so one further unit of time is needed to erase the deficit.