Asymptotic behavior of roots of an equation involving exponential and logarithm Prelude
This Post is a continuation of this Original Post. The original problem asked is:
How many solutions does the following equation have:
$$
a^x = \log_a(x) \,,\quad a \in (0,1) \wedge x \in\mathbb{R}^+_0
$$
And already has been answered (see Claude Leibovici's answer for details).
Observations
I found this problem interesting and I numerically investigated it a bit deeper. This led me to another question. Figures below show that such roots exist and are reals:



And roots can be computed numerically for several values of  $a$:

My observations, so far, are:


*

*All roots must lie in $(0,1)$ because $\log_a(x) > 0 \, \forall x  \in (0,1)$ and $\log_a(x) < 0$ elsewhere, and $a^x > 0 \,\forall x \in \mathbb{R}$;

*Solving this problem involves complex analysis (such as the use of Lambert W function) but the result will stay in the real domain;

*A "divergence" point occurs at $(e^{-e},e^{-1})$;

*Solving the original problem is equivalent to solve (base conversion and Lambert W properties):$$\ln(a) = \frac{W\left(x \ln(x)\right)}{x} = \frac{\ln(x)}{x} \Leftrightarrow a_k = \exp\left[\frac{W_k\left(x \ln(x)\right)}{x}\right]$$

*Roots become triple when $a < e^{-e}$ (dashed black vertical line, as shown by Claude Leibovici in his answer)

*Roots have asymptotic behavior, it can be checked in term of $a(x)$ for two branches:


*

*one root tends to unity as $a\rightarrow 1$: $\lim\limits_{x\rightarrow 1} a = 1$ (green curve rightmost);

*two roots tend to zero as $a\rightarrow 0^+$: $\lim\limits_{x\rightarrow 0^+} a = 0$ (green and orange curves leftmost).



Questions:
My main questions are:


*

*How can I prove that one root tends to unity when base $a\rightarrow 0^+$ (blue curve leftmost)? By taking the limit of the proper branch.

*Is the point 4 correct? Investigating the solution using Wolfram Alpha it seems both Leibovici and my expressions are equivalent. But the first form suffer a huge float arithmetic error with numpy library. Anyway they can be plotted using the latter form:



Side questions are:


*

*How is called the point where branches diverge?

*Do end of the branches also have a specific name?

*Can we say that roots are multiple at the "divergence" point? If so, in what sense are they multiple? Claude Leibovici: Roots are multiple in the sense that three first degrees of Taylor expansion vanishes at $x=e^{-1}$ with $a=e^{-e}$.

*Is the green branch a specific one because it behaves smoothly? 

 A: Probably not answering the questions but this is too long for comments.
Considering the function $$f(x)=a^x-\frac{\log (x)}{\log (a)}$$ its derivatives are
$$f^{(n)}(x)=a^x \log^n(a)+(-1)^n \frac{(n-1)!}{x^n\, \log(a)}$$
The first derivative cancels at two points given by 
$$x_1=\frac{W_0\left(\frac{1}{\log (a)}\right)}{\log (a)}\qquad \text{and}\qquad x_2=\frac{W_{-1}\left(\frac{1}{\log (a)}\right)}{\log (a)}$$ which, in the real domain, exist if $\frac{1}{\log (a)}\geq -\frac 1 e$ that is to say if $a \leq e^{-e}$. If this is the case, $f(x_1)<0$ and $f(x_2)>0$ which explains the three roots.
What is interesting is to look at what happens when $a = e^{-e}$. For this value, the solution of $f(x)=0$ is unique $x=\frac 1e$. At this point, the second derivative is also zero and the Taylor expansion is
$$\frac{e^2}{6}  \left(x-\frac{1}{e}\right)^3-\frac{5e^3}{24} 
   \left(x-\frac{1}{e}\right)^4+O\left(\left(x-\frac{1}{e}\right)^5\right)$$ which makes that, at ths point, $x=\frac 1e$ is a triple root of the equation.
On another side, we could also solve the equation for $a$ and its solutions are given by
$$a_1=\left(\frac{x \log (x)}{W_{0}(x \log (x))}\right)^{\frac{1}{x}}\qquad \text{and}\qquad a_2=\left(\frac{x \log (x)}{W_{-1}(x \log (x))}\right)^{\frac{1}{x}}$$ which do exist if $x \leq \frac 1e$. These two functions are worth to be plotted.
When $x \to 1$ the expansion of $a_1$ is
$$a_1=1+(x-1)-(x-1)^2+\frac{1}{2} (x-1)^3+O\left((x-1)^4\right)$$ and using series reversion
$$x= 1+(a_1-1)+O\left((a_1-1)^2\right)$$ making that if $x\to 0 \implies a_1 \to 0$.
