Consider real roots to the equation $$ \frac{2}{1- \varepsilon x^{2}} = e^{x} $$ as $\varepsilon \to 0$. From the monotonicity of $e^{x}$ together with qualitative properties of $\frac{2}{1-\varepsilon x^{2}}$, we conclude that there exist two roots $x_1, x_2$ with $x_1 \sim \ln(2), x_2 \sim \varepsilon^{-1/2}$.
I have no problem setting up an iteration scheme for $x_1$ to extract higher order terms, but have been stuck on the asymptotic expansion of $x_2$, as I have been unable to tame the singularity of the rational function.
We can rescale through $x \equiv \varepsilon^{-1/2} y$, hoping to develop an equation of the form
$$ y = 1 + \phi(y) \hspace{10 mm} \phi(y)=o(1)$$
Upon rescaling, our equation becomes
$$ \frac{2}{1- y^{2}} = e^{\varepsilon^{-1/2} y} $$
for which I still get stuck and am unable to extract an equation of the form above.
I would appreciate any help with tackling this problem, and the use of asymptotics for transcendental equations in general!
Why not consider that we look for the zeros of function $$f(x)=(1-\epsilon x^2)e^x-2$$ and use series expansion followed by series reversion.
This would give with $L=\log(2)$, $$x_1=L+\frac{\epsilon L^2 }{1-(2L+L^2) \epsilon}+\cdots \tag 1$$ and with $K=\epsilon^{-\frac 12}$, $$x_2=K- K e^{-K}+\cdots \tag 2$$
which are equivalent to the first iterate of Newton method. For sure, we could do better adding more terms in the initial expansion.
Playing with various values of $\epsilon$, these seem to be more than decent.
For $x_2$, a better approximation could be $$x_2=K+\frac{2 K}{2 K-2 e^K+1}$$