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(A) For $PROPERTY \ 1$,we check $\lim_{h \rightarrow 0} \frac{f(h)-f(0)}{\sqrt{|h|}}$:$(2) \ f(x)=x^{2/3}, f(0)=0$. $\lim_{h \rightarrow 0} \frac{h^{2

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$(2, 4)$

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(A) For $PROPERTY \ 1$,we check $\lim_{h \rightarrow 0} \frac{f(h)-f(0)}{\sqrt{|h|}}$:$(2) \ f(x)=x^{2/3}, f(0)=0$. $\lim_{h \rightarrow 0} \frac{h^{2/3}-0}{|h|^{1/2}} = \lim_{h \rightarrow 0} \frac{|h|^{2/3}}{|h|^{1/2}} = \lim_{h \rightarrow 0} |h|^{1/6} = 0$. This exists and is finite. So,$(2)$ is correct.$(4) \ f(x)=|x|, f(0)=0$. $\lim_{h \rightarrow 0} \frac{|h|-0}{|h|^{1/2}} = \lim_{h \rightarrow 0} |h|^{1/2} = 0$. This exists and is finite. So,$(4)$ is correct.For $PROPERTY \ 2$,we check $\lim_{h \rightarrow 0} \frac{f(h)-f(0)}{h^2}$:$(1) \ f(x)=x|x|, f(0)=0$. $\lim_{h \rightarrow 0} \frac{h|h|}{h^2} = \lim_{h \rightarrow 0} \frac{|h|}{h}$. The $RHL = 1$ and $LHL = -1$. The limit does not exist. So,$(1)$ is incorrect.$(3) \ f(x)=\sin x, f(0)=0$. $\lim_{h \rightarrow 0} \frac{\sin h}{h^2} = \lim_{h \rightarrow 0} \frac{\sin h}{h} \cdot \frac{1}{h} = 1 \cdot \infty = \infty$. The limit does not exist. So,$(3)$ is incorrect.Thus,only $(2)$ and $(4)$ are correct.

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