Find the value of $\lim_{x\to 0} \frac{\sin{\{x\}}}{\{x\}}$ 
Find the value(If exists) of $$\lim_{x\to 0} \frac{\sin{\{x\}}}{\{x\}}$$ (where $\{x\}=x-\lfloor x \rfloor $,$\lfloor x \rfloor $ means the greatest integer less than or equal to $x$.

My mind tells me that it doesn't exists but I haven't any prove.I tried to prove it but failed.Please help me.
 A: $$\lim_{x\to0^-} \frac{\sin\{x\}}{\{x\}}=\lim_{x\to0^-} \frac{\sin(x+1)}{(x+1)}=\sin(1)$$ and $$\lim_{x\to0^+} \frac{\sin\{x\}}{\{x\}}=\lim_{x\to0^+} \frac{\sin(x)}{x}=1 \neq \sin(1)$$ The left and right hand limits are not equal, therefore the limit does not exist.
A: Hint:
If $x \in [-1,0)$ then $\lfloor x \rfloor=-1$.
A: To prove the limit does not exist. 
take $$a_=\frac{1}{n} ,b_n=\frac{-1}{n}$$note that $$a_n ,b_n \to 0  \text{ when } n\to \infty$$   now put $a_n,b_n \to x$
$$\lim_{x\to 0} \frac{\sin{\{x\}}}{\{x\}}=\lim_{n\to \infty} \frac{\sin{\{a_n\}}}{\{a_n\}}\tag{1}$$
$$\lim_{x\to 0} \frac{\sin{\{x\}}}{\{x\}}=\lim_{n\to \infty} \frac{\sin{\{b_n\}}}{\{b_n\}}\tag{2}$$
note that 
for (1)$$\lim_{x\to 0} \frac{\sin{\{x\}}}{\{x\}}=\lim_{n\to \infty} \frac{\sin{\{a_n\}}}{\{a_n\}}=\lim_{n\to \infty} \frac{\frac{1}{n}}{\frac{1}{n}} \to 1$$
for (2) 
$$\lim_{x\to 0} \frac{\sin{\{x\}}}{\{x\}}=\lim_{n\to \infty} \frac{\sin{\{b_n\}}}{\{b_n\}}=\\\lim_{n\to \infty} \frac{\sin{\{\frac{-1}{n}\}}}{\{\frac{-1}{n}\}}=\\
\lim_{n\to \infty} \frac{\sin(1-\frac{1}{n})}{(1-\frac{1}{n})}\to sin 1$$ so :
the limit does not exist because $$a_n,b_n \to 0 \\but\\\lim_{n\to \infty}f(a_n) \neq \lim_{n\to \infty}f(b_n)$$
