For $m,n\in\mathbb{Z}$ with $1\leq m\leq n$ consider the polynomial $$ p_{m,n}(x)=x^{m+n+1}-x^{m+n}-1. $$
I am trying to prove that the largest positive root of $p_{m,n}$ tends to $1$ as $m\to\infty$ or $n\to\infty$. And I would like to know if my argument is okay.
The strategy of the proof is the following.
(1.) On $[0,1]$ we have that $p_{m,n}<0$.
(2.) On $(1,\infty)$, $p_{m,n}$ is strictly increasing and convex, hence there is exactly one positive root $\theta_{m,n}$ which then automatically is the largest positive root.
(3.)
$$ p_{m,n}\left(1+\frac{\log(m+n)}{m+n}\right)=\left(1+\frac{\log(m+n)}{m+n}\right)^{m+n}\cdot\frac{\log(m+n)}{m+n}-1 $$
For large $m$ or $n$, we should have $$ \left(1+\frac{\log(m+n)}{m+n}\right)^{m+n}\sim m+n $$
and hence, for large $m$ or $n$, $$ p_{m,n}\left(1+\frac{\log(m+n)}{m+n}\right)\sim\log(m+n)-1>0. $$
Therefore, we have that $$ \theta_{m,n}\in\left(1,1+\frac{\log(m+n)}{m+n}\right) $$
and hence, $\theta_{m,n}\to 1$ as $m\to\infty$ or $n\to\infty$.
