Prove that $\lim_{x\to\infty}\sin(x-\lfloor x\rfloor) $ doesn't exist I need to prove that $\lim_{x \to \infty} {\sin(x-\lfloor x\rfloor)}$ doesn't exist. I wanted to prove it using the definition only so I assumed the limit exists and is equal $L$. I showed that for $L<0$ and $L=0$, $L$ doesn't exist but I can't find the right epsilon in order to prove that for $L>0$. The logic is to find an $x_0$ so that $f(x_0)$ will be greater then L+epsilon or lower then L-epsilon and then You get the limit can't be $L$. Please use my logic and help me find the right epsilon for $L>0$. (The epsilon for $L<0$ was $-L + 0.25$)
 A: First of all note that $0 \leq x - \lfloor x \rfloor < 1$ for all $x$ and hence $0 \leq \sin(x - \lfloor x \rfloor) < \sin 1 < 1$ Therefore the limit $\lim_{x \to \infty}\sin(x - \lfloor x \rfloor)$ (if it exists) has to be in the interval $[0, \sin 1]$. If we take any number $L > \sin 1$ then $\epsilon = L - \sin 1$ will do the job of showing that $\lim_{x \to \infty}\sin(x - \lfloor x \rfloor) \neq L$. Similarly if $L < 0$ then any $0 < \epsilon < -L$ will do the job.
Next let $L \in [0, \sin 1]$ where we have a chance that $\lim_{x \to \infty}\sin(x - \lfloor x \rfloor)$ might be equal to $L$. If $L = 0$ then we can safely choose $\epsilon = (\sin 1)/2$ and if $0 < L \leq \sin 1$ then we can choose $\epsilon = L/2$. Idea is that we have to choose value of $\epsilon $ such that that there should be some part (say an interval of non-zero length) of interval $[0, \sin 1]$ which has no points common with interval $(L - \epsilon, L + \epsilon)$. If $\epsilon$ has such a value then we know that as $ x \to \infty$ the function $(x - \lfloor x \rfloor)$ will take all values in $[0, 1)$ and hence $\sin(x - \lfloor x \rfloor)$ will take all values in $[0, \sin 1)$ and hence there will be some values of this function outside the interval $(L - \epsilon, L + \epsilon)$ and thus $L$ will not be the limit.
