Question

Let $F:R\to R$ be a function. We say that $f$ has

PROPERTY $1$ if $\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{\sqrt{\left|h\right|}}$ exists and is fine

PROPERTY $2$ if $\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h^2}$ exists and is fine

Then which of the following options is/are correct?

• A. $f\left(x\right)=x\left|x\right|$ has $\text{PROPERTY 2}$
• B. $F\left(x\right)=x^{2/3}$ has $\text{PROPERTY 1}$
• C. $f\left(x\right)=\sin x$ has $\text{PROPERTY 2}$
• D. $f\left(x\right)=\left|x\right|$ has $\text{PROPERTY 1}$

Answer 1

Answer: (B), (D)

$f\left(x\right)=\left|x\right|$ has $\text{PROPERTY 1}$

Explanation for correct option(s)

Option B : $F\left(x\right)=x^{2/3}$ has PROPERTY $1$

Consider the given equation as,

$\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{\sqrt{\left|h\right|}}$

If $F\left(x\right)=x^{2/3}$

$$\begin{gathered} \underset{h\to +0}{\lim }\frac{h^{2/3}-0}{\sqrt{\left|h\right|}}=\underset{h\to +0}{\lim }\frac{h^{2/3}}{\sqrt{\left|h\right|}} \\ =0 \\ \underset{h\to -0}{\lim }\frac{-h^{2/3}+0}{\sqrt{\left|h\right|}}=\underset{h\to -0}{\lim }\frac{-h^{2/3}}{\sqrt{\left|h\right|}} \\ =0 \end{gathered}$$

Both values are same so that the limit dos exist

Option D: $f\left(x\right)=\left|x\right|$ has PROPERTY $1$

If $f\left(x\right)=\left|x\right|$

$$\begin{gathered} \underset{h\to +0}{\lim }\frac{\left|h\right|-0}{\sqrt{\left|h\right|}}=\underset{h\to +0}{\lim }\frac{\left|h\right|}{\sqrt{\left|h\right|}} \\ =0 \\ \underset{h\to -0}{\lim }\frac{-\left|h\right|+0}{\sqrt{\left|h\right|}}=\underset{h\to -0}{\lim }\frac{-\left|h\right|}{\sqrt{\left|h\right|}} \\ =0 \end{gathered}$$

Both values are same so that the limit dose exist

Explanation for incorrect option(s)

Option A: $f\left(x\right)=x\left|x\right|$ has PROPERTY $2$

Consider the given equation as,

$\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h^2}$

if $f\left(x\right)=x\left|x\right|$

$$\begin{gathered} \underset{h\to +0}{\lim }\frac{h\left|h\right|-0}{h^2}=\underset{h\to +0}{\lim }\frac{h\left|h\right|}{h^2} \\ =\underset{h\to +0}{\lim }\frac{h^2}{h^2} \\ =1 \end{gathered}$$

$$\begin{gathered} \underset{h\to -0}{\lim }\frac{h\left|h\right|-0}{h^2}=\underset{h\to -0}{\lim }\frac{-h\left|h\right|}{h^2} \\ =\underset{h\to -0}{\lim }\frac{-h^2}{h^2} \\ =-1 \end{gathered}$$

Both values are not same so that the limit dose not exist

Option C: $f\left(x\right)=\sin x$ has PROPERTY $2$

Consider the given equation as,

$\underset{h\to 0}{\lim }\frac{f\left(h\right)-f\left(0\right)}{h^2}$

if $f\left(x\right)=\sin x$

$$\begin{gathered} \underset{h\to +0}{\lim }\frac{\sin \left(h\right)-\sin \left(0\right)}{h^2}=\underset{h\to +0}{\lim }\frac{\sin \left(h\right)}{h^2} \\ =\underset{h\to +0}{\lim }\frac{\sin \left(h\right)}{h}\frac{1}{h} \end{gathered}$$

$$\begin{gathered} \underset{h\to -0}{\lim }\frac{-\sin \left(h\right)+\sin \left(0\right)}{h^2}=\underset{h\to -0}{\lim }\frac{-\sin \left(h\right)}{h^2} \\ =\underset{h\to -0}{\lim }\frac{-\sin \left(h\right)}{h}\frac{1}{h} \end{gathered}$$

Both values are not same so that the limit dose not exist

Therefore, the correct answers are option (B) and (D).