Evaluate $\lim\limits_{n\to\infty}\left(\sqrt [n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac {a_n}{b_n}}\ \right)$ 
Consider two sequences $(a_n)_{n\ge 1}$ and $(b_n)_{n\ge 1}$ of
  positive real numbers such that 
$$\lim_{n\to\infty} \frac {a_{n+1}}{n^2\cdot a_n}=x>0$$
and 
$$\lim_{n\to\infty} \frac {b_{n+1}}{n\cdot b_n}=y>0$$
Evaluate:
$$\lim_{n\to\infty}\left(\sqrt [n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac  {a_n}{b_n}}\ \right)$$

My thoughts: Intuitively $\dfrac{a_{n+1}}{a_n} \sim xn^2$ and $\dfrac{b_{n+1}}{b_n} \sim yn$, so $a_n \sim x \sqrt[n]{(n!)^2}$ and $b_n \sim y\sqrt[n]{n!}$. 
Since $\lim\limits_{n\to \infty} \dfrac{n}{\sqrt[n]{n!}} = e$, we should get:
$$\sqrt[n+1]{\frac{a_{n+1}}{b_{n+1}}} \sim \frac{x}{y}\cdot \sqrt[n+1]{(n+1)!} \sim\frac{x}{y}\cdot \frac{n+1}{e}$$
and the final answer should be $\dfrac{x}{ye}$. But how do we write this with sound arguments?

Some observations. Define
$$
c_n : = \frac{1}{n!}\frac{a_n}{b_n}.
$$
Then by the assumptions on $a_n$ and $b_n$, we have
$$
\mathop {\lim }\limits_{n \to  + \infty } \frac{c_{n + 1}}{c_n} = \frac{x}{y}\quad \text{ and hence }\quad \mathop {\lim }\limits_{n \to  + \infty } \sqrt[n]{c_n} = \frac{x}{y}.
$$
Note that
$$
\sqrt[n]{\frac{a_n}{b_n}} = \sqrt[n]{n!c_n}.
$$
By Stirling's formula
$$
\sqrt[n]{{n!}} = \frac{n}{e} + \frac{{\log (2\pi n)}}{{2e}} + \mathcal{O}\left( {\frac{1}{n}} \right).
$$
If it was true that
$$
\sqrt[n]{{c_n }} = \frac{x}{y} +\frac{K}{n}+ o\left( {\frac{1}{n}} \right)
$$
then we would have
$$
\sqrt[{n + 1}]{{(n + 1)!c_{n + 1} }} - \sqrt[n]{{n!c_n }} = \frac{x}{y}\frac{1}{e} + \mathcal{O}\left( {\frac{1}{n}} \right),
$$
meaning that the limit is indeed $\frac{x}{y}\frac{1}{e}$.
In any case,
$$
\mathop {\lim \inf }\limits_{n \to  + \infty } \left( {\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}} \right) = \mathop {\lim \inf }\limits_{n \to  + \infty } \frac{{\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}}}{{n + 1 - n}} \le \mathop {\lim }\limits_{n \to  + \infty } \frac{{\sqrt[n]{{n!c_n }}}}{n} = \frac{x}{{ye}} \\ \le \mathop {\lim \sup }\limits_{n \to  + \infty } \frac{{\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}}}{{n + 1 - n}} = \mathop {\lim \sup }\limits_{n \to  + \infty } \left( {\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}} \right).
$$
Thus, if the limit in question exists, it must be $\frac{x}{y}\frac{1}{e}$.
