 Power-law crossover in a singular perturbation

For $ϵ>0$, the two terms in $y=x^{-1/ϵ}+x$ are equal at $x=1$. Their ratio is larger than one for $0<x<1$ and smaller than one for $x>1$. As $ϵ\to 0^{+}$ it tends to infinity at fixed $x<1$ and to zero at fixed $x>1$. Comparable contributions require $|\log x|=O(ϵ)$, equivalently $x=1+O(ϵ)$.

To resolve this region exactly, introduce the inner variable $X=\log x/ϵ$. Then $y=e^{-X}+e^{ϵX}\sim e^{-X}+1$ for bounded $X$. At fixed $x>1$, the first term $\exp [-(\log x)/ϵ]$ is smaller than every positive integer power of $ϵ$, so a regular perturbation in powers of $ϵ$ misses it. At fixed $x<1$, that same term diverges. This explains why a power series approximation valid on one side need not be a uniform approximation across the crossover.