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

Answer 1

Answer: (A), (B)

$P-1:$
$\underset{h\to 0}{\lim }$ $\frac{f\left(h\right)-f\left(0\right)}{\sqrt{|h|}}=$ exist and finite
$\left(B\right)f\left(x\right)=x^{2/3},$ $\underset{h\to 0}{\lim }$ $\frac{h^{2/3}-0}{\sqrt{|h|}}=$ $\underset{h\to 0}{\lim }$ $\frac{{|h|}^{2/3}}{\sqrt{|h|}}=0$
$\left(D\right)f\left(x\right)=\left|x\right|,$ $\underset{h\to 0}{\lim }$ $\frac{|h|-0}{\sqrt{|h|}}\Rightarrow$ $\underset{h\to 0}{\lim }$ $\sqrt{\left|h\right|}=0$
$P-2:$
$\underset{h\to 0}{\lim }$ $\frac{f\left(h\right)-f\left(0\right)}{h^2}=$ exist and finite
$\left(A\right)ƒ\left(x\right)=x|x|,$ $\underset{h\to 0}{\lim }$ $\frac{h|h|-0}{h^2}=$ $\left[\begin{array}{ccc}RHL& =\underset{h\to 0}{\lim }& \frac{h^2}{h^2}=1\\ & & \\ LHL& =\underset{h\to 0}{\lim }& \frac{-h^2}{h^2}=-1\end{array}\right]$
$\left(C\right)ƒ\left(x\right)=sinx$ $\underset{h\to 0}{\lim }$ $\frac{sinh-0}{h^2}=DNE$