# Solving $\frac{x}{x-1} + e^{(x-1)/\epsilon} - \epsilon = 0.$

I have this eq. $$\frac{x}{x-1} + e^{(x-1)/\epsilon} - \epsilon = 0.$$

I need to derive an asymptotic expansion for any bounded roots $x(\epsilon)$ this equation might have, to find the coefficient of the $n$th (‘general’) term and show that asymptotic expansion converges for all $\epsilon$ in a neighborhood of zero. Does this asymptotic expansion converge to $x(\epsilon)$ for $\epsilon\neq 0$?

For $\epsilon$ near zero, the term $e^{(x-1)/\epsilon}$ will be extremely small unless $x \gt 1$. So if we ignore it, we have $\epsilon=1+\frac{1}{x-1}$ or $x=\frac{\epsilon}{\epsilon -1}$, safely less than 1. Numerically you can iterate this, plugging this $x$ into the exponential term and solving again. Convergence will be quite rapid. To prove there is a root, you should be able to find points of opposite signs on either side of this $x$.