When does $\displaystyle\lim_{n\to\infty}x_n^{1/n} = \alpha$ imply $\displaystyle\lim_{n\to\infty}\frac{x_n}{\alpha^n}$ exists? Let $x_n$ be a sequence of numbers satisfying 
$$
0 < x_n \leq x_{n-1} \leq \ldots \leq x_0 = 1, \quad n \in \mathbb{N}
$$
as well as
$$
    \lim_{n \to \infty} x_n
=   0, \quad \text{and } \quad
    \lim_{n\to \infty} \frac{x_n}{x_{n-1}}
=   \alpha 
\in (0,1).
$$
From this it follows that 
$$
  \lim_{n \to \infty} x_n^{1/n}
= \alpha;
$$
my question is, under what conditions (the milder the better) can one conclude that the limit
$$
  L
= \lim_{n \to \infty} \frac{x_n}{\alpha^n}
$$
exists? I don't think it is true in general, but there are extra properties that the $x_n$ satisfy which may be germaine and sufficient to secure the result; for the moment though, I'll keep the question unfettered by potential red herrings. 
Some thoughts which may or may not be useful:
the sequence $\phi_n := \frac{x_n}{\alpha^n}$ can be shown to be submultiplicative, i.e. 
$$
     \phi_{n+m}
\leq \phi_n \phi_m, \quad n,m \in \mathbb{N},
$$
and since $x_n$ decreases from $1$ to $0$, there is a point at which it "crosses" the line $y = \alpha$, after which point $\frac{x_n}{\alpha} < 1$.
EDIT
Etienne gave a nice answer. It did not apply directly to my specific situation, but I believe it handles the original question nicely. The additional details of my specific problem are unlikely to be searched for - at least much less likely than the original question. So I'll accept the answer and leave it at that!
Thanks!
 A: Here are simple conditions under which $\frac{x_n}{\alpha^n}$ does converge, without even assuming that $(x_n)$ is decreasing or that $\frac{x_n}{x_{n-1}}\to\alpha$:
(i) If $\frac{x_n}{x_{n-1}}\leq\alpha$ for all $n$ then $\frac{x_n}{\alpha^n}$ converges (in $\mathbb R$).
(ii) If $\frac{x_n}{x_{n-1}}\geq\alpha$ for all $n$, then $\frac{x_n}{\alpha^n}$ converges in $\mathbb R\cup\{+\infty\}$.
To prove this, set $u_n:=-\log(x_n)$. If (i) holds, then $u_{n}-u_{n-1}\geq \beta:=-\log(\alpha)$ for all $n$. So the sequence $v_n:=u_n-n\beta$ is nondecreasing. Hence $v_n$ has a limit in $\mathbb R\cup\{+\infty\}$, so that $\frac{x_n}{\alpha^n}=e^{-v_n}$ has a limit in $\mathbb R$. If (ii) holds, then $v_n$ is nonincreasing, so it has a limit in $\mathbb R\cup\{-\infty\}$ and hence $\frac{x_n}{\alpha^n}$ has a limit in $\mathbb R\cup\{ +\infty\}$.
However, as you guessed, the result is not true in general. 
Consider for example the sequence defined by $x_0=1$ and the "recurrence relation"
$$x_{n}=x_{n-1}\left(\alpha-\alpha_n\right)$$
where $(\alpha_n)_{n\geq 1}$ is a sequence of real numbers tending to $0$ such that $\alpha_n<\alpha$ for all $n$, $1+\inf_{n\geq 1}  \alpha_n>\alpha$ and, moreover, the series $\sum\log\left(1-\frac{\alpha_n}{\alpha}\right)$ is not convergent and does not diverge to $-\infty$.  For an explicit example, one may take $\alpha_n:=\frac{(-1)^n}{\sqrt n}\delta$, where $0<\delta <\alpha$ and $1-\delta>\alpha$; or (much simpler), any sequence of negative numbers $\alpha_n$ such that $1+\inf_{n\geq 1}\alpha_n>\alpha$ and $\sum_1^\infty\alpha_n=-\infty$.
Then $0<x_{n+1}\leq cx_n$ for all $n$, where $c=\alpha-\inf_{n\geq 1}\alpha_n<1$; so $(x_n)$ is decreasing and $x_n\to 0$. Moreover, $\frac{x_{n+1}}{x_n}\to\alpha$. On the other hand, we have (for $n\geq 1$)
$$\frac{x_n}{\alpha^n}=\frac1{\alpha^n}\prod_{k=1}^n \left(\alpha-\alpha_k\right)=\prod_{k=1}^n \left(1-\frac{\alpha_k}{\alpha}\right)\, .$$
Taking the logarithms (which is possible because $1-\frac{\alpha_k}{\alpha}>0$, this gives
$$\log\left(\frac{x_n}{\alpha^n}\right)=\sum_{k=1}^n\log\left(1-\frac{\alpha_k}{\alpha}\right)\, ; $$
so $\log\left(\frac{x_n}{\alpha^n}\right)$ does not converge in $\mathbb R$ and does not tend to $-\infty$ either, and hence $\frac{x_n}{\alpha^n}$ does not converge. 
Edit. If one takes $\alpha_n:=a(1-a^{\sqrt{n}-\sqrt{n+1}})$, one gets Clin's example. If one takes $\alpha_n=-\frac\alpha{n}$ (assuming $\alpha<1/2$), one gets the extremely simple $$x_n= (n+1)\alpha^n\, .$$ 
Note that $\alpha<1/2$ is necessary only to ensure that $(x_n)$ is decreasing. If one forgets this requirement, one may take 
$$x_n = C_n \,\alpha^n$$
where $(C_n)$ is any sequence of positive numbers such that $C_n\to\infty$ and $\frac{C_{n+1}}{C_n}\to 1$. This sequence will be decreasing with $x_0=1$ if $C_0=1$ and $\frac{C_{n+1}}{C_n}<\frac1\alpha$ for all $n$.
A: We can not conclude that $\lim  \frac{x_n}{a^n}$ exists. Consider 
$$x_n=a^{n-\sqrt n}$$
then $\dfrac{x_n}{a^n}=a^{-\sqrt n}$, so $\lim  \frac{x_n}{a^n}=+\infty$
