# If $\,x>1$, then $\lim\limits_{n\rightarrow\infty}\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}=x$. [closed]

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 $z$.

## closed as off-topic by anomaly, Rory Daulton, Thomas, Yagna Patel, PragabhavaJan 22 '16 at 22:35

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• $\frac{\left\lfloor x^{n+1} \right\rfloor}{\left\lfloor x^n \right\rfloor}$ is nearly $\frac{x^{n+1}}{x^n}$. – Hurkyl Mar 10 '14 at 22:05
• (-1) "This question does not show any research effort" – TMM Mar 17 '14 at 21:23

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 $\frac{\lfloor x^{n+1}\rfloor}{\lfloor x^n\rfloor}\to x$.
Hint we have $$y-1\le\left\lfloor y \right\rfloor\le y$$ then use the squeeze theorem.