Properties of Limits for Absolute and Trigonometric Functions

To determine which of the given options are correct, we evaluate the given limit definitions for each option.

Given Definitions:

• PROPERTY 1: $\lim_{h \to 0} \frac{f(h) - f(0)}{\sqrt{|h|}}$ exists and is finite.
• PROPERTY 2: $\lim_{h \to 0} \frac{f(h) - f(0)}{h^2}$ exists and is finite.

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Option (A): $f(x) = |x|$ has PROPERTY 1

For $f(x) = |x|$, we have $f(0) = 0$. Checking Property 1:

$$\lim_{h \to 0} \frac{f(h) - f(0)}{\sqrt{|h|}} = \lim_{h \to 0} \frac{|h| - 0}{\sqrt{|h|}} = \lim_{h \to 0} \sqrt{|h|} = 0$$

Since the limit exists and is finite ($0 \in \mathbb{R}$), $f(x) = |x|$ satisfies PROPERTY 1. Thus, Option (A) is correct.

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Option (B): $f(x) = x^{2/3}$ has PROPERTY 1

For $f(x) = x^{2/3} = (|h|^2)^{1/3} = |h|^{2/3}$, we have $f(0) = 0$. Checking Property 1:

$$\lim_{h \to 0} \frac{f(h) - f(0)}{\sqrt{|h|}} = \lim_{h \to 0} \frac{|h|^{2/3} - 0}{|h|^{1/2}} = \lim_{h \to 0} |h|^{\frac{2}{3} - \frac{1}{2}} = \lim_{h \to 0} |h|^{1/6} = 0$$

Since the limit exists and is finite ($0 \in \mathbb{R}$), $f(x) = x^{2/3}$ satisfies PROPERTY 1. Thus, Option (B) is correct.

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Option (C): $f(x) = x|x|$ has PROPERTY 2

For $f(x) = x|x|$, we have $f(0) = 0$. Checking Property 2:

$$\lim_{h \to 0} \frac{f(h) - f(0)}{h^2} = \lim_{h \to 0} \frac{h|h|}{h^2} = \lim_{h \to 0} \frac{|h|}{h}$$

Evaluating the one-sided limits:

• Right-hand limit (RHL): $\lim_{h \to 0^+} \frac{h}{h} = 1$
• Left-hand limit (LHL): $\lim_{h \to 0^-} \frac{-h}{h} = -1$

Since $\text{RHL} \neq \text{LHL}$, the limit does not exist. Hence, $f(x) = x|x|$ does not satisfy PROPERTY 2. Thus, Option (C) is incorrect.

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Option (D): $f(x) = \sin x$ has PROPERTY 2

For $f(x) = \sin x$, we have $f(0) = 0$. Checking Property 2:

$$\lim_{h \to 0} \frac{f(h) - f(0)}{h^2} = \lim_{h \to 0} \frac{\sin h}{h^2} = \lim_{h \to 0} \left(\frac{\sin h}{h} \cdot \frac{1}{h}\right)$$

Since $\lim_{h \to 0} \frac{\sin h}{h} = 1$ and $\lim_{h \to 0} \frac{1}{h}$ does not exist (tends to $\pm \infty$), the limit does not exist as a finite value. Hence, $f(x) = \sin x$ does not satisfy PROPERTY 2. Thus, Option (D) is incorrect.

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Conclusion:

The correct options are A and B.