JEE Main 2024 — 8 April, Shift 1 — Question 30

$\lim _{x \rightarrow 0} 2\left(\frac{1-\left(1-\frac{x^{2}}{2!}\right)\left(1-\frac{4 \mathrm{x}^{2}}{2!}\right)\left(1-\frac{9 \mathrm{x}^{2}}{2!}\right) \cdots \cdot\left(1-\frac{100 \mathrm{x}^{2}}{2!}\right)}{\mathrm{x}^{2}}\right)$ By expansion

$\lim _{\mathrm{x} \rightarrow 0} \frac{2\left(1-\left(1-\frac{\mathrm{x}^{2}}{2}\right)\right)\left(1-\frac{1}{2} \cdot \frac{4 \mathrm{x}^{2}}{2}\right)\left(1-\frac{1}{3} \cdot \frac{9 \mathrm{x}^{2}}{2}\right) \cdots . .\left(1-\frac{1}{10} \cdot \frac{100 \mathrm{x}^{2}}{2}\right)}{\mathrm{x}^{2}}$

$\lim _{x \rightarrow 0} 2\left(\frac{1-\left(1-\frac{x^{2}}{2}\right)\left(1-\frac{2 x^{2}}{2}\right)\left(1-\frac{3 x^{2}}{2}\right) \ldots\left(1-\frac{10 \mathrm{x}^{2}}{2}\right)}{x^{2}}\right)$

$\lim _{x \rightarrow 0} \frac{2\left(1-1+x^{2}\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots . .+\frac{10}{2}\right)\right)}{x^{2}}$

$2\left(\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\ldots .+\frac{10}{2}\right)$

$1+2+\ldots \ldots+10=\frac{10 \times 11}{2}=55$