Continuity and Differentiability
Limits Using Taylor Series
JEE Advanced 2022 Paper 2
Grade 12
Question:
If $\beta = \lim_{x \to 0} \frac{e^{x^3} - (1 - x^3)^{1/3} + ((1 - x^2)^{1/2} - 1) \sin x}{x \sin^2 x}$, then the value of $\beta$ is _____.
Step-by-Step Solution
Key Concept: Expand each piece using Taylor series around $x = 0$: $e^{x^3} \approx 1 + x^3$; $(1 - x^3)^{1/3} \approx 1 - x^3/3$; $(1 - x^2)^{1/2} \approx 1 - x^2/2$; $\sin x \approx x$; $\sin^2 x \approx x^2$. Numerator $\approx (x^3 + x^3/3) + (-x^2/2)(x) = \frac{4x^3}{3} - \frac{x^3}{2} = \frac{5x^3}{6}$. Denominator $\approx x^3$. $\beta = 5/6$.
The detailed step-by-step mathematical proof is available inside the Mathbee app workspace.
Correct Answer: 5/6