Prove
$$ \displaystyle \lim_{x \to 0^+}\frac{0}{\sin x} = 0 $$
I will show that:
$$\forall \xi_0 > 0, \exists \delta > 0, \forall x \in R, 0 < |x - 0| < \delta: |f(x) - 0| < \xi_0$$
Let $\delta = \frac{\pi}{2} \Rightarrow 0 < x < \frac{\pi}{2} \Rightarrow 0 < \sin x < 1 \Rightarrow \frac{0}{\sin x} = 0 < \xi_0$
Therefore: $$ \displaystyle \lim_{x \to 0^+}\frac{0}{\sin x} = 0 $$
$\frac{0}{\sin(x)} = 0$ for all $x \neq k\pi$ for $k \in\mathbb{Z}$, hence, for any $0<|x-0| < \pi$, we have $|\frac{0}{\sin(x)} - 0| = |0|=0 < \epsilon$.