JEE Advanced 2022 Paper 2 · Q05 · Limits

Question

If $\beta = \displaystyle\lim_{x	o 0}\dfrac{e^{x^{3}} - (1-x^{3})^{1/3} + ((1-x^{2})^{1/2}-1)\sin x}{x\,\sin^{2}x}$, then the value of $6\beta$ is _____.

SolveKar AI · Accepted Answer

Use Taylor expansions: $e^{x^3}=1+x^3+\dfrac{x^6}{2}+\cdots$; $(1-x^3)^{1/3}=1-\dfrac{x^3}{3}-\dfrac{x^6}{9}+\cdots$; $(1-x^2)^{1/2}-1=-\dfrac{x^2}{2}-\dfrac{x^4}{8}+\cdots$; $\sin x = x-\dfrac{x^3}{6}+\cdots$. Numerator: $(e^{x^3}-(1-x^3)^{1/3}) = \dfrac{4x^3}{3}+O(x^6)$; $((1-x^2)^{1/2}-1)\sin x = -\dfrac{x^3}{2}-\dfrac{x^5}{8}+\cdots+\dfrac{x^5}{12}\cdots = -\dfrac{x^3}{2}+O(x^5)$. Sum $=\dfrac{4x^3}{3}-\dfrac{x^3}{2}+O(x^5)=\dfrac{5x^3}{6}+O(x^5)$. Denominator $x\sin^2 x = x\cdot x^2 + O(x^5)= x^3$. So $\beta=\dfrac{5}{6}$, and $6\beta=5$.

JEE Advanced 2022 Paper 2 · Q05 · Limits

Solution