Let $1\leq a\leq b$ with $a,b\in\mathbb{N}$. Moreover, let $c:=a+b$.
Consider the polynomial $$ f_{c }(x)=x^{c+1}-x^{c}-1. $$ By $x_{c}$ denote the largest positive root of $f_{c}$.
It can be shown that, for large $c$, $$ x_c\sim1+\frac{\log c}{c}~~~(*) $$ and hence $\lim_{c\to\infty}x_{c }=1$.
Note $f_{c}$ is the characteristic polynomial of the $(c+1)\times (c+1)$-matrix $A_{c}$ with entries $$ a_{1,1}=1, \quad a_{c+1,1}=1 $$ and $a_{i,j}=1$ on the right side-diagonal and 0 elsewhere.
For example, if $c=2$, then $$ A_{2}=\begin{pmatrix}1 & 1 & 0\\0 & 0 & 1\\1 & 0 & 0\end{pmatrix}. $$
Question
I wonder: Does there exists a „scaling exponent“ $\alpha=\alpha(c)$ such that
(1) $x_{c}^{\alpha(c)}$ is the largest root of the characteristic polynomial of the matrix $A_{c}^{\alpha(c)}$ and
(2) the limit $\lim_{c\to\infty}x_c^{\alpha(c)}\neq 1$ exists?
At least for exponents $\alpha(c)\in\mathbb{N}$, ad (1) should be satisfied.
Using $(*)$, one should have that, for large $c$, $$ x_c^{\alpha(c)}\sim\left(1+\frac{\log c}{c}\right)^{\alpha(c)} $$ So, maybe it is useful to consider $$ \lim_{c\to\infty}\left(1+\frac{\log c}{c}\right)^{\alpha(c)} $$ Here, I dont know how to continue.
If $c$ is even, then $A_c$ has no $\leq 0$ eigenvalues. Then, for any real $\alpha(c)$, you can define ${A_c}^{\alpha(c)}$ as a real matrix with $\rho({A_c}^{\alpha(c)})={x_c}^{\alpha(c)}$.
It suffices to choose $\alpha(c)=\dfrac{c}{\log(c)}$. Indeed $\log({x_c}^{\alpha(c)})\sim \alpha(c)\dfrac{\log(c)}{c}=1$ and ${x_c}^{\alpha(c)}$ tends to $e$.
Yet, you obtain the same result if you choose $\alpha(c)=Int(\dfrac{c}{\log(c)})$ ("Int" denotes the integer part). Moreover, for such an $\alpha(c)$, the result works also for odd $c$.
EDIT. I forgot to say that $x_c$ is the root of $f_c$ with maximum modulus.