If $$ \beta = \lim_{x \rightarrow 0} \frac{e^{x^3} - (1-x^3)^{\frac{1}{3}} + ((1-x^2)^{\frac{1}{2}} - 1)\sin x}{x\sin^2 x} $$ then the value of $6\beta$ is ____.

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Limits XII

β = x \to 0 lim \frac{e ^{x^{3}} - ( 1 - x ^{3} ) ^{\frac{1}{3}} + (( 1 - x ^{2} ) ^{\frac{1}{2}} - 1 ) sin x}{x sin ^{2} x}

then the value of

6 β

is ____.

Answer: 5

Rewrite the limit:

β = x \to 0 lim \frac{e ^{x^{3}} - ( 1 - x ^{3} ) ^{1/3}}{x ^{3}} + x \to 0 lim \frac{(( 1 - x ^{2} ) ^{1/2} - 1 ) sin x}{x ^{3}}

Using Taylor expansions:

First limit:

x \to 0 lim \frac{( 1 + x ^{3} ) - ( 1 - \frac{x ^{3}}{3} )}{x ^{3}} = x \to 0 lim \frac{\frac{4 x ^{3}}{3}}{x ^{3}} = \frac{4}{3}

Second limit:

x \to 0 lim \frac{( 1 - \frac{x ^{2}}{2} - 1 ) \cdot x}{x ^{3}} = x \to 0 lim \frac{- \frac{x ^{3}}{2}}{x ^{3}} = - \frac{1}{2}

Therefore:

β = \frac{4}{3} - \frac{1}{2} = \frac{5}{6}

6 β = 5

JEE Advanced · 2022 · 28 Aug 2022 · Paper 2 · Integer answer