Let f : mathbb R rightarrow mathbb R be a function. We say that f has PROPERTY 1 if lim h rightarrow 0 frac f( h )- f (0) sqrt mid hmid exists and is finite, and PROPERTY 2 if lim h rightarrow 0 frac f(h)-f(0) h 2 exists and if finite. Then, which of the following options is/are correct?

f ( x )= x 2 / 3 has PROPERTY 1

Mathematics · Limits, Continuity and Differentiability

JEE Advanced 2019 — Paper 2 — Question 27

Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be a function. We say that f has

PROPERTY 1 if $\lim _{h \rightarrow 0} \frac{f(\mathrm{~h})-\mathrm{f}(0)}{\sqrt{\mid \mathrm{h\mid}}}$ exists and is finite, and

PROPERTY 2 if $\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h^{2}}$ exists and if finite.

Then, which of the following options is/are correct?

(A) $f(x)=x|x|$ $=\lim _{h \rightarrow 0} \frac{h|h|}{h^{2}}$

$\lim _{\mathrm{h} \rightarrow 0} \frac{|\mathrm{~h}|}{\mathrm{h}}$ does not exist.

(B) $\mathrm{f}(\mathrm{x})=\sin \mathrm{x}$

$\lim _{h \rightarrow 0} \frac{\sinh -0}{h^{2}}$ does not exist.

\lim _{\mathrm{h} \rightarrow 0} \frac{|\mathrm{~h}|-0}{\sqrt{|\mathrm{~h}|}}=0

(D) $f(x)=x^{2 / 3}$

\lim _{h \rightarrow 0} \frac{h^{2 / 3}}{\sqrt{|h|}}=0

Answer key and solution verified before publishing.

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