Show that there is no limit exists for $\lim_{x \rightarrow \infty}\sin(x-\lfloor x \rfloor)$

Question

Question as above is required to use $ε-δ$ definition. I do assume the limit exists for the function $f(x)=\sin(x-\lfloor(x)\rfloor$) is $L$ when $x$ approaches infinity. Since $f(x)$ is a well-defined function, and $0\leqslant x-\lfloor x \rfloor<1$, indicates that $f(x)$ is bounded at the interval of $[0,\sin1)$. I then discuss the cases for: (1) $L<0$; (2) $L>\sin1$, but no idea comes out to show the case for $0\leqslant L \leqslant \sin 1$ by choosing a suitable point $x_0$ and $\delta$ for the case. Any good idea to simplify my method of proving this?

Answer 1

A beginning:

We want to show for an arbitrary $L$ that $f(x)$ does not converge to $L$. Since $f$ is not a constant function, there are $a$ and $b$ such that $f(a)\ne f(b)$. So $f(a)$ and $f(b)$ cannot both equal $L$, so at least one of them is differs from $L$. Without loss of generality suppose $f(a)\ne L$. Set $\varepsilon = \frac{|f(a)-L|}{2}$ ...

(and remember that $f(a+k)=f(a)$ for every $k\in\mathbb N$).

Answer 2

This is my solution and can you guys help me to check if there's a problematic? Since the function $f$ is a periodic function with period 1, there are some $x_a$ and $x_b$ such that $f(x_a)=a$ and $f(x_b)=b$, where $a\neq b$. Suppose the function $f$ has a limit of $L$ for $x \rightarrow \infty$. Take $\epsilon=\frac{|a-b|}{4}$, then there exists a positive number $N$ such that $\forall x>N$ such that $|f(x)-L|<\epsilon=\frac{|a-b|}{4}$ holds. However, plugging $x=x_a$ and $x=x_b$ into the expression of limit above gives $L<\frac{3a+b}{4}$ or $L<-\frac{a-5b}{4}$, together with condition $x_a,x _b>R$ due to periodicity of function $f$, obtaining a contradiction. Hence, limit of the function $f$ does not exist for $x$ approaches infinity. I am not pretty sure if my proof is getting in a correct way but it is the best way I could present in this question.