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. 

I will show,
by elementary means that
the root is beteween
$1+\dfrac1{\sqrt{a+1}}$
and
$1+\dfrac1{a+1}$.
Since the root of
$x^{a+2}-x^{a+1}-1
$
is close to $1$,
let
$x = 1+y$.
Then we want
$1
=(1+y)^{a+2}-(1+y)^{a+1}
=(1+y)^{a+1}(1+y-1)
=y(1+y)^{a+1}
$.
Since
$(1+y)^{a+1}
\gt 1+y(a+1)
$,
$1
\gt y(1+y(a+1))
\gt y^2(a+1)
$
so
$y \lt \dfrac1{\sqrt{a+1}}
$.
Since
$(1+1/(a+1))^{a+1}
< e
$,
if
$y < 1/(a+1)
$
then
$(1+y)^{a+1}
< e
$
so that
$y(1+y)^{a+1}
< ey
<\dfrac{e}{\sqrt{a+1}}
< 1
$
for
$a > 8$.
Therefore
$y > \dfrac1{a+1}$.
