I am considering the polynomial $$ p(c,x):=x^{c+1}-x^c-1 $$ By $x_c$ denote the largest positive real root of $p(c,x)$.
From mathematics we know that $x_c\in (1,1+\frac{\log(c)}{c})$ and that, for large $c$, $x_c$ behaves like $1+\frac{\log(c)}{c}$ which means, in mathematical terms, that $x_c\sim 1+\frac{\log(c)}{c}$ as $c\to\infty$ or more formally: $$ \lim_{c\to\infty}\frac{x_c}{1+\frac{\log(c)}{c}}=1. $$
My numerics show that we also seem to have $$ \lim_{c\to\infty}\frac{x_c}{1+\frac{1}{c}}=1. $$
In the following picture, the blue curve is the function $$ f(c)=\frac{x_c}{1+\frac{\log(c)}{c}} $$ and the orange curve is the function $$ g(c)=\frac{x_c}{1+\frac{1}{c}}. $$
It seems that the blue curve reaches the value 1 faster than the orange curve.
Does this mean that it is better to say that $$ x_c\sim 1+\frac{\log(c)}{c}, c\to\infty $$ than to say that $$ x_c\sim 1+\frac{1}{c}, c\to\infty? $$
Note $\frac{1}{c}\leq\frac{\log(c)}{c}$ for large $c$. So I would have expected that $1+\frac{1}{c}$ is a better approximation.
For all you know, $x_c$ could be equal to $1+\frac{1}{c}$, and it could as well be equal to $1+\frac{\log c}{c}$. All of what you said would still hold. Except that obviously the "best" approximation in not the same in both cases. So you should not expect $\frac{1}{c}\leq\frac{\log(c)}{c}$ to tell you anything about which approximation is the best.
Secondly, note that $$\lim_\limits{c\to \infty}\frac{1+\frac{\log(c)}{c}}{1+\frac{1}{c}}=1$$ therefore any sequence equivalent to the numerator is also equivalent to the denominator. Which of these approximations is the "best" depends on the sequence. You can't tell anything more without looking at higher order terms.