limit of implicit sequences In a HS problem, we study the functions 
$$f_n:x \longmapsto (x-n)\ln x - x\ln(x-n)$$ for $n$ natural with $n\ge 5$
First questions just ask the domain, the monotonicity ($f_n$ is defined for $x> n$ and is strictly decreasing on its domain) then by IVT we show that the equation $f_n(x)=0$ has a unique solution $\alpha_n$ that satisifes $n+1 < \alpha_n <n+2$.
From last inequality it's clear that $(\alpha_n)$ diverges to $+\infty$.
The last question asks to prove that $\lim\limits_{n \to +\infty} (\alpha_n - n)=1$ and then to evaluate $\lim\limits_{n \to +\infty} (\alpha_{n+1} - \alpha_n)$.
I'm stuck with this last question.
 A: For the first part note that
$$\frac{\alpha_n-n}{\log(\alpha_n - n)} = \frac{\alpha_n}{\log(\alpha_n)}  $$
Since $\alpha_n \to \infty$, the right hand side diverges to infinity so the left hand side must diverge to infinity as well. Since $1 < \alpha_n - n < 2$ for all $n$, this implies that $\log(\alpha_n-n) \to 0$ i.e. $\alpha_n - n \to 1$.
For the second part
$$ \begin{align} \lim(\alpha_{n+1} - \alpha_n) & = \lim((a_{n+1} - (n+1)) - (\alpha_n - n) + 1) \\ & = \lim(\alpha_{n+1} - (n+1)) - \lim(\alpha_n - n) + 1 \\ & = 1 - 1 + 1 \\ & = 1 \end{align}$$
A: This could be a stupid answer.
You look for the zero of
$$f_n=(x-n)\log( x) - x\log(x-n)$$ and you showed that the root is between $n+1$ and $n+2$. So,let $x=n+1+\epsilon$ to make
$$f_n=(1+\epsilon ) \log (n+1+\epsilon)-(n+1+\epsilon) \log (1+\epsilon)$$ Expand it as a Taylor series for infinitely large values of $n$ to get
$$f_n=-n \log (1+\epsilon)-(1+\epsilon)\log \left(\frac{1+\epsilon }{n}\right)+O\left(\frac{1}{n}\right)$$ So, if $n\to \infty$, $\epsilon\to 0$
