Let $f(x)=\frac{x\ln x}{1+x}$,evaluate $\lim _{m\to f\left(x_0\right)+}\frac{x_1+x_2-2x_0}{m+x_0}$. Let $f(x)=\frac{x\ln x}{1+x}$,suppose $x_0$is the extreme point of $f$.For $m\in \mathbb{R} $,it's easy to see  that if $m>f(x_0)=-x_0$,then the equation $f(x)=m$ has exactly two real roots $x_1,x_2$,$x_1<x_0<x_2$.The question is how to evaluate the limit
\begin{align*}
\lim _{m\to f\left(x_0\right)+}\frac{x_1+x_2-2x_0}{m+x_0}？
\end{align*}
To this problem,we can see that if $m\to f(x_0)+$,then $x_1\to x_0^-$,and $x_2\to x_0^+$.By Taylor's formula we have
$$f(x_1)-f(x_0)=f'(x_0)(x_1-x_0)+\frac12f''(\xi)(x_1-x_0)^2=\frac12f''(\xi)(x_1-x_0)^2,$$
where $\xi \in (x_1,x_0)$.So
$$m+x_0\sim \frac{f''(x_0)}{2}(x_1-x_0)^2.$$
But I can't handle $x_1+x_2-2x_0$.If we can expand $x_1+x_2-2x_0$ as  terms of power of $(x_1-x_0)$,then it must be
$$x_1+x_2-2x_0\sim k(x_1-x_0)^2.$$
But how can I get it?Maybe there are other ways.
 A: The essence of this question is to expand the roots (technically functional inverses for the multivalued function presented above) in a series around the minimum, where they coincide. As demonstrated by @ClaudeLeibovici above, a second-order approximation is required to evaluate the limit, and we will find that for a general function $f(z)$. Suppose that at $z=z_0$ the function attains a critical point with $f(z_0)=\zeta,f'(z_0)=0, f''(z_0)\neq 0$. Denote the two local branches of the inverse with $f^{-1}_{\pm}(z),f^{-1}_{\pm}(\zeta)=z_0$, where the plus/minus branch indicates the root that in some disk around the critical point obeys $(f^{-1}_+(z)>z_0)/(f^{-1}_-(z)<z_0)$.
How do we find successive monomial approximations to these? It is not a priori clear what the form is close to the critical point, so we try to evaluate the first-order limit
$$L_{1,\pm}=\lim_{z\to \zeta}\frac{f_{\pm}^{-1}(z)-z_0}{(z-\zeta)^a}$$
where $a\in \mathbb{R}$ is to be defined such that the limit above exists and is finite. Perform the change of variables $z=f(t)$:
$$L_{1,\pm}=\lim_{t\to z_0^{\pm}}\frac{t-z_0}{(f(t)-\zeta)^a}$$
Given the Taylor series expansion $f(t)-\zeta=\frac{f''(z_0)}{2}(t-z_0)^2+\mathcal{O}((t-z_0)^3)$ around $z_0$ we readily deduce that for this limit to be finite the exponent needs to take on the value $a=1/2$ and the limit evaluates to
$$L_{1,\pm}=\pm\sqrt{\frac{2}{f''(z_0)}}$$
Following the same thought process we can continue to compute succesive approximations by subtracting the last approximation from the series and dividing by an appropriate power of $z-\zeta$. For the second-order approximation it looks like this:
$$L_{2,\pm}=\lim_{z\to \zeta}\frac{f^{-1}_{\pm}(z)-z_0-(\pm)\sqrt{\frac{2}{f''(z_0)}}\sqrt{z-\zeta}}{(z-\zeta)^a}=\lim_{z\to z_0^{\pm}}\frac{t-z_0-(\pm)\sqrt{\frac{2}{f''(z_0)}}\sqrt{f(t)-\zeta}}{(f(t)-\zeta)^a}$$
We expand the numerator to second order to find that
$$t-z_0-\pm\sqrt{\frac{2}{f''(z_0)}(f(t)-\zeta)}=\frac{f'''(z_0)}{6f''(z_0)}(t-z_0)^2+\mathcal{O}((t-z_0)^{3})$$
Thus $a=1$ and the limit evaluates to
$$L_{2,\pm}=-\frac{f'''(z_0)}{3(f''(z_0))^2}$$
We have computed enough terms to calculate the desired limit. In this notation it can be written
$$\mathcal{L}=\lim_{m\to f(x_0)}\frac{f^{-1}_{+}(m)+f^{-1}_{-}(m)-2x_0}{m-f(x_0)}=2L_{2,\pm}=-\frac{2}{3}\frac{f'''(x_0)}{(f''(x_0))^2}\approx 1.40924 $$
This calculation hints at the fact that there is a generalization of Lagrange's inversion theorem around a critical point. I have not been able to find a general formula for the coefficients of the branches of the inverse, but I think it is possible to obtain such a formula.
It also turns out that there is an exact solution to this equation! Set in the original equation $z=\ln x-m$ which transforms it to the form $ze^z=m e^{-m}$ which admits two solutions
$$x_1(m)=\frac{m}{W_0(me^{-m})}~,~ x_2(m)=\frac{m}{W_{-1}(me^{-m})}$$
However this form is not particularly useful for obtaining expansions.
A: I do not expect any bounty for this, just wanted to share my thoughts.
Note that in this problem we have implicitly defined functions $x_1(m)$ and $x_2(m)$.
Here is a graph displaying $x_1(m)$ and $x_2(m)$. (Their graphs lie in the second quadrant and are traced out in black as we use the slider to move $m$ closer to $-x_0$.)
The limit we seek,  $\displaystyle L:= \lim_{m \rightarrow f(x_0)^+} \frac{x_1(m)+x_2(m)-2x_0}{m+x_0}$,  is of the form $\frac{0}{0}$ so we can try to apply L'Hopital's Rule.  Unfortunately, doing so yields $\displaystyle L= \lim_{m \rightarrow f(x_0)^+} \frac{x_1'(m) + x_2'(m)}{1}$ which evaluates to $-\infty + \infty$.
Note that we can also split the limit into two parts to obtain  $$\displaystyle L= \lim_{m \rightarrow f(x_0)^+} \frac{x_2(m)-x_0}{m - (-x_0)} +\lim_{m \rightarrow f(x_0)^+} \frac{x_1(m)-x_0}{m - (-x_0)}.$$
The difference quotients that appear in the above expression are the slopes of secant lines from $(-x_0,x_0)$ to $(m, x_2(m))$ and from $(-x_0,x_0)$ to $(m, x_1(m))$, respectively.  The first represents a positive slope while the second a negative one.  Perhaps these slopes cancel in a controlled way as $m \rightarrow -x_0$ so the limit of their sum exists (as a real number), but I expect they do not.
