Let $P_n=X^n+X-1$ be a polynomial with only one positive real root $x_n$.
The sequence $(x_n)$ is obviously well defined and it's not hard to find that $\lim\limits_{n \to +\infty} x_n=1$
In fact $(x_n)$ is an increasing bounded sequence. Let $l$ be its limit. We obviously have $l \le 1$.
As $P_n$ is increasing on $\mathbb{R_+}$,
$0=x_n^n+x_n-1\le l^n+l-1$ so the limit when $n \to \infty$ satisfies $l-1\ge 0$.
So $l=1$.
Then I need to find an equivalent sequence to $x_n-1$.
What I did:
We have $n=\frac{\ln(1-x_n)}{\ln x_n}$.
I therefore tried to find an equivalent of the form $n^{\alpha}(\ln n)^{\beta}$ and after some trials I found that $-\frac{\ln n}{n}$ works.
Since
$\displaystyle\frac{n}{\ln n} (1-x_n)=\displaystyle\frac{\frac{\ln(1-x_n)}{\ln x_n} (1-x_n)}{\ln\frac{\ln(1-x_n)}{\ln x_n}}$ we only need to show that $\displaystyle\lim\limits_{n \to +\infty}\frac{\ln(1-x_n)}{\ln\frac{\ln(1-x_n)}{\ln x_n}}=-1$
because $\lim\limits_{n \to +\infty}\frac{(1-x_n)}{\ln x_n}=-1$ (limit of the tangency point of $\ln$ at $1$).
we can show that $\displaystyle\lim\limits_{x \to 1^{-}}\frac{\ln(1-x)}{\ln\frac{\ln(1-x)}{\ln x}}=-1$
To simplify it, I noticed that both numerator and denominator tend to $\pm \infty$ when $x\to 1^-$.
So we can apply L'Hôpital's Rule and then :
$\frac{d}{dx} \ln(1-x)=\frac1{x-1}$
and $\frac{d}{dx}\left( \ln\frac{\ln(1-x)}{\ln x} \right)=\frac{x \ln x +(1-x)\ln{(1-x)}}{x(x-1) \ln x \ln(1-x)}$
the limit becomes $\frac{x \ln x +(1-x)\ln{(1-x)}}{x\ln x \ln(1-x)}=\frac 1 {\ln(1-x)}+\frac{(1-x)}{x\ln x} \to_{x \to 1^-} 0+(-1)=-1$.
Finally we have that $\frac{n}{\ln n} (1-x_n) \to -1$ and then $x_n-1 \sim -\frac{\ln n}{n}$
My questions:
- Is this correct ?
- Is there a better way to proceed ? Maybe with the MVT ?
Thanks in advance.
