Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 336 / 1 / a / Solution

The divide-and-conquer asymptotic expansion separates the $x=O(a)$ endpoint region from the $x=O(1)$ bulk, whose expansions individually contain terms that are nonuniform in the other region. Here the recombined answer can also be checked exactly. Put $y=\sqrt{a+x}$; thenAs $a\to 0^{+}$,Multiplication givesThe nonanalytic $\sqrt{a}$ term is the contribution that a naive fixed-$x$ expansion misses at the endpoint.