Question

$\underset{x\to 1}{lim}\frac{(1-x)(1-x^2)...(1-x^{2n})}{{\left\{\right(1-x\left)\right(1-x^2)...(1-x^n\left)\right\}}^2},n\in N$, equals

• A. ${}^{2n}{P}_n$
• B. ${}^{2n}{C}_n$
• C. $\left(2n\right)!$
• D. None of these

Answer 1

Answer: (B)

${}^{2n}{C}_n$

$\underset{x\to 1}{lim}\frac{(1-x)(1-x^2)\dots (1-x^{2n})}{{\left\{\right(1-x\left)\right(1-x^2)\dots (1-x^n\left)\right\}}^2}$
$=\underset{x\to 1}{lim}\frac{\left(\frac{1-x}{1-x}\right)\left(\frac{1-x^2}{1-x}\right)\dots \left(\frac{1-x^{2n}}{1-x}\right)}{{(\left(\frac{1-x}{1-x}\right)\left(\frac{1-x^2}{1-x}\right)\dots \left(\frac{1-x^n}{1-x}\right))}^2}$
$=\frac{1\times 2\times 3\times \dots \left(2n\right)}{(1\times 2\times 3\dots n{)}^2}=\frac{\left(2n\right)!}{n!n!}={}^{2n}{C}_n$
Hence, option 'B' is correct.