$$\lim_{x\to 1}{\left(\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{({(1-x)(1-x^2)\cdots(1-x^n)})^2}\right)}$$ i am a 12th class student and this question was given by our mathematics teacher a year ago. this is part of JEE EXAM syllabus , i tried in this way.. $$\lim_{x\to 1}{\left(\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{({(1-x)(1-x^2)\cdots(1-x^n)})^2}\right)}$$ $$ =\lim_{x\to 1}{\left(\frac{(1-x^{n+1})(1-x^{n+2})\cdots(1-x^{2n})}{{(1-x)(1-x^2)\cdots(1-x^n)}}\right)}$$ then knowing that $$\lim_{x\to 1}{(\frac{1-x^n}{1-x})}=n$$ hence
$$\lim_{x\to 1}{\left(\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{({(1-x)(1-x^2)\cdots(1-x^n)})^2}\right)} = \binom{2n}{n}$$ Is this right? Is there any other way to do it?
We have that
$$(1-x^k)=(1-x)(\overbrace{1+x+\ldots x^{k-1}}^{k\,terms})$$
therefore
$$\lim_{x\to 1}{\left(\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{({(1-x)(1-x^2)\cdots(1-x^n)})^2}\right)}=$$$$=\lim_{x\to 1} {\left(\frac{(1)(1+x)\cdots(1+x+\ldots+x^{2n-1})}{({(1)(1+x)\cdots(1+x+\ldots+x^{n-1})})^2}\right)}=\frac{(2n)!}{(n!)^2}=\binom{2n}{n}$$