Let $f: R \to R$ be a function. We say that $f$ has PROPERTY 1: If $\lim_{h \to 0} \frac{f(h) - f(0)}{\sqrt{|h|}}$ exists and is finite, and PROPERTY 2: If $\lim_{h \to 0} \frac{f(h) - f(0)}{h^{2}}$ exists and is finite Then which of the following options is/are correct?

$\text{Property 1: } \lim_{h\to 0}\dfrac{f(h)-f(0)}{\sqrt{|h|}} \text{ exists \& is finite}$ $\text{Property 2: } \lim_{h\to 0}\dfrac{f(h)-f(0)}{h^{2}} \text{ exists \& is finite}$ $(1)\ f=x^{2/3}:\ \text{P1};\quad (2)\ f=\sin x:\ \text{P2}$ $(3)\ f=|x|:\ \text{P1};\quad (4)\ f=x|x|:\ \text{P2}$ $\text{P1}: \dfrac{f(h)-f(0)}{\sqrt{|h|}} \to L_{1} \in \mathbb{R}$ $\text{P2}: \dfrac{f(h)-f(0)}{h^{2}} \to L_{2} \in \mathbb{R}$ $\text{check each option's ratio as } h\to 0$ $\text{P1} \Rightarrow f(h)-f(0)=L_{1}\sqrt{|h|}+o(\sqrt{|h|})$ $\Rightarrow f \text{ is continuous at } 0$ $\text{P2} \Rightarrow f(h)-f(0)=L_{2}h^{2}+o(h^{2})$ $\Rightarrow f'(0) \text{ exists and } f'(0)=0$ $\text{P2 is much stronger than P1}$ $\text{P2} \Rightarrow \text{P1, but not conversely}$ $f(0)=0,\quad f(h)=|h|^{2/3}$ $\dfrac{f(h)-f(0)}{\sqrt{|h|}}=\dfrac{|h|^{2/3}}{|h|^{1/2}}$ $=|h|^{1/6}\to 0 \quad (\text{finite})$ $\therefore (1)\ \text{is correct} \ \checkmark$ $f=|x|:\quad \dfrac{|h|}{\sqrt{|h|}}=\sqrt{|h|}\to 0$ $\therefore (3)\ \text{is correct} \ \checkmark$ $f=\sin x:\quad \dfrac{f(h)-f(0)}{h^{2}}=\dfrac{\sin h}{h^{2}}$ $\dfrac{\sin h}{h^{2}}\sim \dfrac{1}{h}\to +\infty \ (h\to 0^{+}),\quad -\infty \ (h\to 0^{-})$ $\therefore (2)\ \text{is wrong} \ \times$ $f=x|x|:\quad \dfrac{f(h)-f(0)}{h^{2}}=\dfrac{h|h|}{h^{2}}$ $=\dfrac{|h|}{h}=\operatorname{sgn}h \to +1 \ (h>0),\quad -1 \ (h<0)$ $\therefore (4)\ \text{is wrong} \ \times$ $\boxed{(1)\ \text{and}\ (3)\ \text{are correct}}$