Yet Another Question On Using Basics Limit Arithmetics Is this claim true?
Given $\lim \limits_{n\to \infty}\ a_n=\frac{1}{2}$ Then $\lim \limits_{n\to \infty}\ (a_n - [a_n])=\frac{1}{2}$
I think it's true, but probably I just didn't find the right example to disprove it. 
Thanks a lot.
 A: It depends on what you mean by the symbol $[x]$. If $[x]$ is the floor of $x$ (i.e., the greatest integer less than or equal to $x$), then the limit will be $\frac{1}{2}$. This follows from $\lim \limits_{n\to \infty}\ a_n=\frac{1}{2}$: there exists an $N$ such that for every $n \geq N$, $|a_n - \frac{1}{2}| < \frac{1}{2}$. For these $a_n$, it follows that $[a_n] = 0$, and so $$\lim \limits_{n\to \infty}\ a_n - [a_n] =\frac{1}{2} - 0 = \frac{1}{2}$$
On the other hand, if $[x]$ is the nearest integer to $x$, then the limit does not exist: consider the sequence $a = (0.4,0.6,0.49,0.51,0.499,0.501,\ldots)$. Clearly $a_n \rightarrow \frac{1}{2}$, but $[a_n]$ alternates between $0$ and $1$.
A: Since 
$$\lim_{n \to \infty} a_{n} = \frac{1}{2}$$
there exists $N$ such that for all $n \geq N$, 
$$|a_{n} - \frac{1}{2}| < \frac{1}{4}$$
Then, for $n \geq N$, we have
$$0 \leq a_{n} \leq 1 \Longrightarrow [a_{n}] = 0$$
so that for $n \geq N$,
$$a_{n} - [a_{n}] = a_{n}$$
Now pick $\epsilon > 0$. Suppose that for all $n > M_{\epsilon}$
$$|a_{n} - \frac{1}{2}| < \epsilon$$
Replacing $M_{\epsilon}$ with $\mbox{max}(M_{\epsilon},N)$, we have
$$|(a_{n} - [a_{n}]) - \frac{1}{2} |  = |a_{n} - \frac{1}{2}| < \epsilon$$
which proves that 
$$\lim_{n \to \infty} (a_{n} - [a_{n}]) = \frac{1}{2}$$
A: We want to show that for any $\epsilon>0$, there is an $N$ such that if $n>N$, then $|(a_n-\lfloor a_n\rfloor)-\frac{1}{2}|<\epsilon$.  Here by $\lfloor w\rfloor$ we mean the greatest integer which is $\le w$.
Let $\epsilon'=\min(\epsilon,1/4)$.  By the fact that $\lim_{n\to\infty} a_n=\frac{1}{2}$, there is an $N$ such that if $n>N$ then $|a_n-\frac{1}{2}|<\epsilon'$.  In particular, $|a_n-\frac{1}{2}|<\frac{1}{4}$, and therefore $\lfloor a_n\rfloor=0$. It follows that if $n>N$ then 
$$\left|(a_n-\lfloor a_n\rfloor)-\frac{1}{2}\right|=\left|a_n-\frac{1}{2}\right|<\epsilon.$$
