TITLE: When does $\displaystyle\lim_{n\to\infty}x_n^{1/n} = \alpha$ imply $\displaystyle\lim_{n\to\infty}\frac{x_n}{\alpha^n}$ exists?
QUESTION [5 upvotes]: 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!

REPLY [2 votes]: 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 $\lim  \frac{x_n}{a^n}=+\infty$