<s>[INST] If $\,x>1$, then $\lim\limits_{n\rightarrow\infty}\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}=x$. How can I prove that
$$
\lim_{n\rightarrow\infty}\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}=x,
$$
whenever $x>1$. Here $\left\lfloor \cdot\right\rfloor$ denotes the floor function,
or the integer part function. 
The integer part $\lfloor z\rfloor$ of $z$ is the largest integer, which does not exceed $z$.
Thanks for your answer.
 [/INST] Since $y-1< \lfloor y\rfloor\le y$, for every $y\in\mathbb R$, then
$$
\frac{x^{n+1}-1}{x^{n}}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}< \frac{x^{n+1}}{x^n-1},
$$
and hence
$$
x-\frac{1}{x^n}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}<x+\frac{x}{x^n-1},
$$
or
$$
-\frac{1}{x^n}<\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}-x<\frac{x}{x^n-1}.
$$
Since both
$
-\frac{1}{x^n},\,\frac{x}{x^n-1}\to 0, \quad\text{as}\quad n\to\infty,
$
then $\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}\to x$.
</s>