QJEEAdv22Ma7928: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Ma7928 Solution: If then the value of is ___________.

Given,

$\beta = \mathop {\lim }\limits_{x \to 0} {{{e^{{x^3}}} - {{(1 - {x^3})}^{{1 \over 3}}} + ({{(1 - {x^2})}^{{1 \over 2}}} - 1)\sin x} \over {x{{\sin }^2}x}}$

$= \mathop {\lim }\limits_{x \to 0} {{(1 + {x^3}\, + \,...) - \left( {1 - {{{x^3}} \over 3}\, + \,...} \right) + \left( {\left( {1 - {1 \over 2}{x^2}\, + \,...} \right) - 1} \right)x} \over {x\,.\,{{{{\sin }^2}x} \over {{x^2}}}\,.\,{x^2}}}$

$= \mathop {\lim }\limits_{x \to 0} {{\left( {{x^3} + {{{x^3}} \over 3} - {{{x^3}} \over 2}} \right) + {x^4}(......)} \over {{x^3}}}$

$= \mathop {\lim }\limits_{x \to 0} {{{x^3}\left( {1 + {1 \over 3} - {1 \over 2}} \right)} \over {{x^3}}}$ [Neglecting higher power of x]

$= 1 + {1 \over 3} - {1 \over 2} = {5 \over 6}$

$\therefore$ $6\beta = 6 \times {5 \over 6} = 5$