The behavior of a solution to a parametrized equation Show that the equation
$$x\left(1+\log(\frac{1}{\epsilon \sqrt{x}}\right) = 1, x>0, \epsilon>0$$
has exactly two solutions if $\epsilon>0$ is small enough. In this case, let $x(\epsilon)$ be the smaller solution, show that $x(\epsilon)\to 0$ if $\epsilon\to 0$. 
So far I could solve, just turn the equation into 
$$\epsilon^2 x = e^{2-2/x}$$
where the right hand side has derivative 0 at origin, and convex on $(0,c)$ and then concave on $(c,\infty)$ for some $c$ and $\lim_{x\to \infty} e^{2-2/x} = e^2$. The statements are obvious.
But I wasn't able to solve the last part: show that for any $s>0$, 
$$\epsilon^{-s} x(\epsilon)\to \infty, \epsilon\to 0^+.$$
 A: If $x = 2/w$, the equation can be transformed to 
$$ w = 2 \epsilon^2 e^{-2} \exp(w) $$
which has solutions
$$ w = - W(-2 \epsilon^2 e^{-2})$$
where $W$ is a branch of the Lambert W function.  Now you want $w > 0$, i.e. 
$W(-2 \epsilon^2 e^{-2}) < 0$.  The principal branch and the $-1$ branch of $W(z)$ are both negative for $z \in (-1/e, 0)$, and these are the only branches that can be real for real $z$.  So you have two positive solutions exactly for
$-1/e < -2 \epsilon^2 e^{-2} < 0$, i.e. $0 < \epsilon < \sqrt{e/2}$.
EDIT: As $z \to 0-$, the principal branch of $W(z)$ goes to $0$ and the $-1$ branch goes to $-\infty$, corresponding to $x \to +\infty$ and $x \to 0$ respectively.  In the case of the $-1$ branch, for any $c > 0$ we have $\exp(w) > c w^{1+2/s}$ when
$w$ is large enough, and then
$$ \eqalign{w &= 2 \epsilon^2 e^{-2} \exp(w) > 2 \epsilon^2 e^{-2} c w^{1+2/s}\cr
w^{-2/s} &> 2 \epsilon^2 e^{-2} c \cr
x = 2/w &> 2 (2 e^{-2} c)^{s/2} \epsilon^s \cr
}
$$
But $c$ was arbitrary...
