calculus - Prove that function has no finite limit using $epsilon$ - $delta$ definition

I want to prove using
$$(\exists \varepsilon > 0)(\forall \delta > 0)\exists x(0 < \left| {x - {x_0}} \right| < \delta \Rightarrow \left| {f(x) - L} \right| \ge \varepsilon )$$
That the function $$f(x) = {x \over {\left( {x - \left\lfloor {\sin x} \right\rfloor } \right)}}$$ has no finite limit when $x_0 = 0$
and I can't seem to find the way to start.
I know using the heine method that the function has 1 and 0 limits on both sides, so I'm guessing that if i choose an $\varepsilon=1/2 $ I might be able to show that, but I'm not sure how.

Answer

Hint: You need to show that, for each $L \in \mathbb{R}$, there exists an $\varepsilon > 0$ with the property you mentioned.

Suggestion: Take $\varepsilon = 1/3$. For $L \geq 1/2$, consider points $-\delta < x < 0$; for $L \leq 1/2$, consider points $0 < x < \delta$.