JEE Advanced 2022 Paper 2, Mathematics Section 1 Q5: Methods to Evaluate Limits

Rewrite the denominator as $x{\sin }^{2}x=x^3\cdot \frac{{\sin }^{2}x}{x^2}$, and use $\frac{{\sin }^{2}x}{x^2}\to 1$.

Expand each numerator term to the relevant order in $x$:

$e^{x^3}=1+x^3+\frac{x^6}{2}+\cdots$

$(1-x^3{)}^{1/3}=1-\frac{x^3}{3}-\frac{x^6}{9}-\cdots$

$(1-x^2{)}^{1/2}-1=-\frac{x^2}{2}-\frac{x^4}{8}-\cdots$, and $\sin x=x-\frac{x^3}{6}+\cdots$, so $((1-x^2{)}^{1/2}-1)\sin x=-\frac{x^3}{2}-\frac{x^5}{8}+\cdots$ (leading order $-\frac{x^3}{2}$).

Combine leading $x^3$ coefficients: $1-\left(-\frac{1}{3}\right)+\left(-\frac{1}{2}\right)=1+\frac{1}{3}-\frac{1}{2}=\frac{5}{6}$.

Therefore $\beta =\frac{5}{6}$ and $6\beta =5$.