Question

Let f: R $\to$ ℝ be a function. We say that f has
PROPERTY 1 if $\displaystyle\lim_{h \to 0} \frac{f\left(h\right) - f\left(0\right)}{\sqrt{\left|h\right|}}$ exists and is finite, and
PROPERTY 2 if $\displaystyle\lim_{h \to 0} \frac{f\left(h\right) - f\left(0\right)}{h^{2}}$ 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. $f(x) = sin \, x$ has PROPERTY 2

Answer 1

Answer: (A), (B)

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