Question

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

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

Answer 1

Answer: (A), (B)

The correct options are
A $f\left(x\right)=|x|$ has PROPERTY $1$
B $f\left(x\right)=x^{\frac{2}{3}}$ has PROPERTY $1$

Option 1: $f\left(x\right)=|x|$
Property $1$- $\underset{h\to 0}{lim}\frac{|h|-0}{\sqrt{|h|}}=\underset{h\to 0}{lim}\sqrt{|h|}=0$ (limit exists and finite)
Property $2$-$\underset{h\to 0}{lim}\frac{|h|-0}{|h{|}^2}=\underset{h\to 0}{lim}\frac{1}{h}=\infty$
(limit infinite)

Option 2: $f\left(x\right)=x^{\frac{2}{3}}$
Property $1$- $\underset{h\to 0}{lim}\frac{h^{\frac{2}{3}}-0}{\sqrt{|h|}}$
$\underset{h\to 0}{lim}\frac{h^{\frac{2}{3}}}{h^{\frac{1}{2}}}=0$
$\underset{h\to 0^{-}}{lim}\frac{h^{\frac{2}{3}}}{-h^{\frac{1}{2}}}=0$
(limit exists )
Property $2$-$\underset{h\to 0}{lim}\frac{h^{\frac{2}{3}}-0}{|h{|}^2}=\underset{h\to 0}{lim}\frac{1}{h^{\frac{1}{3}}}=\infty$
(limit infinite )

Option 3: $f\left(x\right)=x|x|$
Property $1$- $\underset{h\to 0}{lim}\frac{h|h|-0}{\sqrt{|h|}}=\underset{h\to 0}{lim}h\sqrt{|h|}=0$ (limit exists and finite)

Property $2$-
$\underset{h\to 0}{lim}\frac{h|h|-0}{|h{|}^2}=\underset{h\to 0}{lim}\frac{|h|}{h}$(limit doesn't exists )

Option 4: $f\left(x\right)=\sin x$
Property $2$-
$\underset{h\to 0}{lim}\frac{\sin h-0}{h^2}=\infty$
(limit infinite )