TITLE: Largest root and asymptotics.
QUESTION [7 upvotes]: I am considering the polynomial
$$
p_c(x)=x^{c+1}-x^c-1.
$$
Its largest root, denoted by $\lambda_c$, is contained in the open interval $(1,1+\frac{\ln c}{c})$.
Making the ansatz $\lambda_c=1+x_c$ for a zero sequence $(x_c)$, I want to express more explicitly what happens with $\lambda_c$ as $c\to\infty$.
Putting the ansatz into the polynomial, I get
$$
(1+x_c)^cx_c=1
$$
which yields, when applying logarithm,
$$
c\ln(1+x_c)+\ln(x_c)=0.
$$
Using the approximation $\ln(1+x_c)=x_c+f(x_c)$ with $f(x_c)=O(x_c^2)$ as $c\to\infty$, this gives
$$
cx_c+cf(x_c)+\ln(x_c)=0.
$$
Exponentiating and multiplicating with the factor $c$ yields
$$
cx_ce^{cx_c}=ce^{-cf(x_c)}
$$
and using Lamberts W-Function, ones gets
$$
x_c=c^{-1}W(ce^{-cf(x_c)}).
$$
Using the large argument approximation
$$
W(x)=\ln x-\ln(\ln(x))+o(1)\text{ as }x\to\infty,
$$
gives me
$$
x_c=\frac{\ln c}{c}-f(x_c)-\frac{\ln(\ln c)}{c}-\frac{\ln(1-\frac{cf(x_c)}{\ln c})}{c}+o\left(\frac{1}{c}\right)\text{ as }c\to\infty
$$


My question is whether this can be also expressed as
    $$
x_c=\frac{\ln c}{c}+o\left(\frac{\ln c}{c}\right)\text{ as }c\to\infty?
$$



For the summands $o\left(\frac{1}{c}\right)$ and $\frac{\ln(\ln c)}{c}$, I can see immediately that they are indeed of order $o\left(\frac{\ln c}{c}\right)$ as $c\to\infty$.
But for the summands $f(x_c)$ and $\frac{\ln(1-\frac{cf(x_c)}{\ln c})}{c}$ this is absolutely not clear to me. 
Edit (due to the comments):
$x_c=o(1)$ and $x_c=O(\ln c/c)$. Both together imply $x_c^2=o(\ln c/c)$, as $c\to\infty$. Moreoever, $f(x_c)=O(x_c^2)$ and $x_c^2=o(\ln c/c)$ imply that $f(x_c)=o(\ln c/c)$ as $c\to\infty$.
Hence, it remains to clarify whether $\frac{\ln(1-\frac{cf(x_c)}{\ln c})}{c}=o(\ln c/c)$ as $c\to\infty$.

REPLY [2 votes]: The claim is equivalent to 
$$F(c)=\dfrac{\ln (1-cf(x_c)/\ln c)}{\ln c} \rightarrow 0,$$
as $c \rightarrow \infty$. Since $|f(x_c)|\leq Kx_c^2$ for some $K>0$ and $x_c<\ln c/c$, we have
$$\ln(1-cf(x_c)/\ln c)\leq \ln (1+cKx_c^2/\ln c) \leq \ln (1+K \ln c/c) $$
and so
$$F(c) \leq \dfrac{\ln (1+K \ln c/c)}{\ln c} \rightarrow 0,$$
since $\ln c /c \rightarrow 0$ as $c \rightarrow \infty$.