JEE Main 2024 — Limits Question with Solution

${\lim }_{x\to 0}2\left(\frac{1-\left(1-\frac{x^2}{2!}\right)\left(1-\frac{4x^2}{2!}\right)\left(1-\frac{9x^2}{2!}\right)\dots .\left(1-\frac{100x^2}{2!}\right)}{x^2}\right)$
By expansion $\begin{gathered} \left.\lim _{x \rightarrow 0} \frac{2\left(1-\left(1-\frac{x^2}{2}\right)\right)\left(1-\frac{1}{2} \cdot \frac{4 x^2}{2}\right)\left(1-\frac{1}{3} \cdot \frac{9 x^2}{2}\right) \ldots . .\left(1-\frac{1}{10} \cdot \frac{100 x^2}{2}\right)}{x^2}\right) \\ \lim _{x \rightarrow 0} 2\left(\frac{\left.1-\left(1-\frac{x^2}{2}\right)\left(1-\frac{2 x^2}{2}\right)\left(1-\frac{3 x^2}{2}\right) \ldots . .\left(1-\frac{10 x^2}{2}\right)\right)}{x^2}\right) \\ \lim _{x \rightarrow 0} \frac{2\left(1-1+x^2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots .+\frac{10}{2}\right)\right)}{x^2} \\ 2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots .+\frac{10}{2}\right) \\ 1+2+\ldots \ldots+10=\frac{10 \times 11}{2}=55 \end{gathered}$