# Find the limit with floor sequence or prove that it doesn't exist

I need to find the limits of these 2 sequences or prove that they don't exist. $$\lim_{x\to\infty}\left(x-5\left\lfloor\frac x5\right\rfloor\right)$$ $$\lim_{x\to\infty}\frac{x-5\left\lfloor \frac x5\right\rfloor}x$$ But I don't know how to get rid of the floor here. I know I should use somehow this inequality. $\frac{x}{5} -1 \leq \left \lfloor \frac{x}{5} \right \rfloor \leq \frac{x}{5}$ But I'm not sure how to do it right. And how to define in any sequence if the limit exist or not? Thank you!

• Hint: The second limit exists and is equal to 0, the first doesn't. To show that the first doesn't exist, consider $x=5n$ and $x=5n+1$ for $n\in\mathbb N$. To show that the second is 0, use your inequalities and estimate the fraction from above and below. Then let $x\to\infty$ on these bounds. – sranthrop Nov 5 '16 at 12:32
• Thank you very much! I just have a question about the 1st one. So if I consider x=5n, the limit of the sequence will be equal to 0 in this case. And if I take x=5n+1, the limit will be equal to 1. So these limits are not the same, and it proves that the limit does not exist, right? – Green Banana Nov 5 '16 at 12:51
• Exactly. Well done :) – sranthrop Nov 5 '16 at 14:50

Here is an answer of the first part. We use the fractional part $\{x\}$ of a real number $x$. $$\{x\}=x-\lfloor x\rfloor\qquad\qquad 0\leq \{x\}<1$$
From the representation (1) we see, the values oscillate in $[0,5)$ when $x$ increases, so that the limit does not exist. In other words: Since the limits of the subsequences \begin{align*} 5\cdot\lim_{{x\to\infty}\atop{x\in 5\mathbb{Z}}}\left\{\frac{x}{5}\right\}=0 \qquad\qquad\text{and}\qquad\qquad 5\cdot\lim_{{x\to\infty}\atop{x\in \frac{5}{2}\mathbb{Z}}}\left\{\frac{x}{5}\right\}=\frac{5}{2}\\ \end{align*} are different, the limit (1) does not exist.