Question

$\underset{x\to -1}{lim}\frac{(1+x)(1-x^2)(1+x^3)(1-x^4)....(1-x^{4n})}{{\left[\right(1+x\left)\right(1-x^2\left)\right(1+x^3\left)\right(1+x^4).......(1-x^{2n}\left)\right]}^2}\text{is equal to:}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$\underset{x\to -1}{lim}\frac{(1+x)(1-x^2)(1+x^3)(1-x^4)....(1-x^{4n})}{{\left[\right(1+x\left)\right(1-x^2\left)\right(1+x^3\left)\right(1+x^4).......(1-x^{2n}\left)\right]}^2}\underset{x\to -1}{lim}\frac{1+x^{2n+1}}{1+x}\times \frac{1-x^{2n+2}}{1-x^2}\times ......\times ....\frac{1-x^{4n}}{1-x^{2n}}\underset{x\to -1}{lim}\frac{x^{2n+1}-(-1{)}^{2n+1}}{x-(-1)}\times \frac{x^{2n+2}-(-1{)}^{2n+2}}{x^2-(-1{)}^2}\times ......\times ....\frac{x^{4n}-(-1{)}^{4n}}{x^{2n}-(-1{)}^{2n}}=\frac{2n+1}{1}.\frac{2n+2}{2}.\frac{2n+3}{3}.\frac{2n+4}{4}......\frac{4n}{2n}{=}^{4n}{C}_{2n}$