# Is there such a thing as $\lim_\limits{x\to\infty}\text{frac}(x)$?

Is there such a thing as $$\lim_{x\to\infty}\text{frac}(x)$$

What would be the reasoning in trying to prove its existence or non-existence?

• Can you give us an example of what you mean by the fractional part of $x$? – DMcMor Jul 11 '17 at 15:07
• By fractional part do you mean the decimal expansion of x? – Ollie Jul 11 '17 at 15:07
• yes. like for 1,5 the fractional part is 0.5, for 20,56 its 0.56 – Alexandra Jul 11 '17 at 15:08
• This limit does not exist. You can show that given any $L$, $N$, and $\epsilon$ there exists an $x>N$ such that $|L-\{x\}|>\epsilon$. This shows that $L$ cannot be the limit of the fractional part of $x$. – Callus Jul 11 '17 at 15:08
• Consider the following increasing sequences for $x$: $1, 2, 3, 4, ...$ and $1.9, 2.9, 3.9, 4.9, ...$. What does $\text{frac}(x)$ converge to in both cases? – Tob Ernack Jul 11 '17 at 15:19

As noted in the comments, the limit does not exist. For clarification, here $\text{frac}(x)$ is the fractional part of $x$, which for positive $x$ is $\text{frac}(x) = x - \lfloor x \rfloor$. Plotted, this function looks like a 'sawtooth' and oscillates with period 1 and amplitude 1, so we would graphically not expect the limit to exist. There are two other ways to see this.
There are two increasing sequences that tend to infinity: $x_n = n$ and $y_n = n + 0.9$. For the first sequence $\lim_{n\rightarrow\infty} \text{frac}(x_n) = 0$ and for the second sequence $\lim_{n\rightarrow\infty} \text{frac}(y_n) = 0.9$, thus the limit $\lim_{x\rightarrow\infty} \text{frac}(x)$ does not exist.
The other way to see this is from the definition of this limit. For any $L$, there exists an $\epsilon$ such that for every $N$, there exists an $x>N$ such that $|L - \text{frac}(x)| > \epsilon$. (When $L \neq 0$, take $\epsilon = \frac{|L|}{2}$ and $x = \lceil N \rceil + 1$. When $L = 0$, take $\epsilon = \frac{1}{2}$ and $x = \lceil N \rceil + 0.9$.)