KVPY Sample Paper KVPY Stream-SX Model Paper-26

Let \[g\left( x \right)=\int_{0}^{{{\left| x \right|}^{3/4}}}{{{t}^{2/3}}}\sin \frac{1}{t}dt,\] for all real x, then \[_{x\to 0}^{\lim }\frac{g\left( x \right)}{x}\] is equal to

A) \[\infty \]

B) \[-\infty \]

C) 0

D) \[\frac{3}{4}\]

Correct Answer: C

Solution :

We have, \[g\left( x \right)=\int_{0}^{\left| x \right|3/4}{{{\left( t \right)}^{2/3}}\sin \left( \frac{1}{t} \right)dt}\]

\[g'\left( x \right)={{\left| x \right|}^{\frac{1}{2}}}\sin \frac{1}{{{\left| x \right|}^{3/4}}}_{x\to 0}^{\lim }\frac{g\left( x \right)}{x}\]

\[_{x\to 0}^{\lim }\frac{g'\left( x \right)}{1}=_{x\to 0}^{\lim }{{\left| x \right|}^{1/2}}\sin \frac{1}{{{\left| x \right|}^{3/4}}}=0\]