Using the expansion for $\arctan(t)$ as $t\to +\infty$ in the form $$\arctan(t) = \frac{\pi}{2}-\frac{1}{t}+\frac{1}{3t^3}+\cdots$$ we get, for fixed $x>0$ and positive $d\to 0$: $$ f_d(x) \sim \frac{\pi(x^2+d)}{4d^{3/2}} =: C_d(x^2+d) $$ and further (that's where the magic cancellation happens) $$g_d(x):=f_d(x)-C_d(x^2+d)\sim \frac{x}{2d}-\frac{(x^2+d)}{2d^{3/2}}\left( \frac{\sqrt{d}}{x}-\frac{d^{3/2}}{3x^3} \right) \\ = \frac{d^2-2dx^2}{6dx^3}\to -\frac{1}{3x}$$
There's nothing deep happening here, except the necessity to keep track of the constant $C$ as the parameter varies (and also of the intervals definition for maximal solutions, which hasn't been checked above).