Find $\lim\limits_{x\rightarrow0}\frac{\sin\{x\}}{\{x\}}$

Question

Here {$x$} denotes the fractional part of x

When $x\rightarrow0^+$, $\{x\}=x$. (Since $x$ will be a very small number example: $0.000000000001,$ it will not have an integer part, and the fractional part will be equal to the number).

When $x\rightarrow0^-$, $\{x\}=1+x$

So $\displaystyle\lim\limits_{x\rightarrow0^+}\frac{\sin\{x\}}{\{x\}}$ simplifies to $\displaystyle\lim\limits_{x\rightarrow0^+}\frac{\sin x}{x}=1$

And $\displaystyle\lim\limits_{x\rightarrow0^-}\frac{\sin\{x\}}{\{x\}}$ simplifies to $\displaystyle\lim\limits_{x\rightarrow0^-}\frac{\sin (x+1)}{x+1}=\sin1$

So it's discontinuous

Edit: Made corrections as suggested in the comments

Answer 1

For $x \in [0,1)$ we have $\{x\} = x - \lfloor x\rfloor = x-0 = x$, but for $x \in [-1,0)$ we have $\{x\} = x- \lfloor x \rfloor = x - (-1) = x+1$.