Largest root as exponent goes to $+\infty$ Let $a\geq 1$ and consider 
$$
x^{a+2}-x^{a+1}-1.
$$
I am interested to see what is the largest root of this polynomial as $a\to +\infty$.
In order to find a root, we surely have to have
$$
x^{a+2}-x^{a+1}=x^{a+1}(x-1)=1.
$$
Hence, I guess we have to look for which $x$ we have that
$$
x^{a+1}(x-1)\to 1\text{ as }a\to+\infty.
$$
Intuitively, if $x$ tends to some value larger than $1$ for $a\to\infty$, the whole thing should diverge. On the other side, if $x$ tends to some value smaller than $1$, then the whole expression should converge to $0$. Hence I guess that $x\to 1$ as $a\to\infty$ in order to get a root. 

If we set
$$ p_a(x) = x^{a+2}-x^{a+1}-1 $$
we may easily see that $p_a(x)$ is negative on $[0,1]$, increasing and convex on $[1,+\infty)$, so the largest real root is in a right neighbourhood of $x=1$. We may also notice that:
$$ p_a\left(1+\frac{\log(a+1)}{a+1}\right) = \frac{\log(a+1)}{a+1}\left(1+\frac{\log(a+1)}{a+1}\right)^{a+1}-1>0 $$
by Bernoulli's inequality, hence the largest root of $p_a$ is between $1$ and $1+\frac{\log(a+1)}{a+1}$.
A more effective localization can be achieved by performing one step of Newton's method with starting point $x=1+\frac{\log(a+1)}{a+1}$.
