Asymptotic behavior of the solution to an equation Let $c\in(0,1)$  be a constant and let $k$ be a positive odd integer, and let $a(k)$ denote the value of $a$ that satisfies the equation $$(1-a)^kk\sqrt{a}=c$$.  As $k\rightarrow\infty$, what can we say about the asymptotic behavior of $a(k)$?  Obviously, $a(k)$ must approach zero, but at what rate?
 A: If we divide through by $k$ the equation becomes
$$
(1-a)^k \sqrt{a} = c/k.
$$
If we suppose that $a = o(1/k)$ then $(1-a)^k \sim 1$, so we get $\sqrt{a} \sim c/k$ and thus $a \sim c^2/k^2$.  This is the smaller of the two positive solutions for odd $k$.
To find the larger solution, take logs of both sides of the equation to get
$$
k \log(1-a) + \tfrac{1}{2}\log a = \log c - \log k. \tag{$*$}
$$
Since $a \to 0$ we have $\log(1-a) \approx -a$, so this equation is analogous to something like
$$
-ka + \tfrac{1}{2}\log a \asymp -\log k.
$$
We can find two solutions to this equation by letting one or the other of the terms on the left-hand side balance with the right-hand side; either


*

*$\log a \asymp -2\log k$, which is the solution $a \sim c^2/k^2$ we found earlier, or

*$ka \asymp \log k$, which is the larger solution.
Note that in the second case we also have a balance between the two terms on the left, $-ka \asymp \log a$, so we won't get the correct leading order term by just supposing that $ka \sim \log k$.  We need to also consider the contribution from the $\log a$ term.
Let's formalize this to explicitly calculate some terms of the asymptotic for $a$.
If we suppose $a \sim b \log k/k$, where $b$ is some constant to be determined, then
$$
\log(1-a) = -a + O(a^2) = -a + o(1/k)
$$
and
$$
\log a = -\log k + \log\log k + \log b + o(1),
$$
so upon substituting these into $(*)$ we obtain
$$
-ka - \tfrac{1}{2} \log k + \tfrac{1}{2}\log\log k + \tfrac{1}{2}\log b + o(1) = -\log k + \log c
$$
or, rearranging,
$$
a = \frac{\log k}{2k} + \frac{\log\log k}{2k} + \frac{\log(\sqrt{b}/c)}{k} + o\left(\frac{1}{k}\right).
$$
We assumed $a \sim b\log k/k$, so from this we can see that $b = 1/2$, and thus
$$
a = \frac{\log k}{2k} + \frac{\log\log k}{2k} - \frac{\log(c\sqrt{2})}{k} + o\left(\frac{1}{k}\right).
$$
Of course this is pretty non-rigorous, but it should be relatively straightforward to justify the error bounds along the way.
