<s>[INST] Calculation of $ \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots(1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $ Calculation of $\displaystyle \lim_{x\rightarrow 1}\frac{(1-x)\cdot(1-x^2)\cdot(1-x^3)\cdots (1-x^{2n})}{\{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)\}^2} = $
My Trial After simplification, we get $$\displaystyle \lim_{x\rightarrow 1}\frac{(1-x^{n+1})\cdot(1-x^{n+2})\cdot(1-x^{n+3})\cdots(1-x^{2n})}{(1-x)\cdot(1-x^2)\cdot (1-x^3)\cdots(1-x^n)}$$
Now I did not understand How can I solve after that, Help me
Thanks
 [/INST] We can use the identity: $$1 - x^m = (1-x)(1 + x + \cdots+ x^{m-1})$$
in every single factor, there. All of those $1-x$ will cancel, and we'll be left with $$\lim_{x \to 1} \frac{(1+x+\cdots+ x^n)(1+x+\cdots+ x^{n+1})\cdots (1+x+\cdots +x^{2n - 1})}{(1+x)(1+x+x^2)\cdots(1+x+x^2+\cdots +x^{n-1})} = \frac{(n+1)(n+2)\cdots 2n}{1\cdot 2 \cdot 3 \cdots n}$$
To put this nicely, see that $$\frac{(n+1)(n+2)\cdots 2n}{1\cdot 2 \cdot 3 \cdots n} = \frac{1\cdot 2 \cdot 3 \cdots n \cdot (n+1)\cdot (n+2)\cdots 2n}{ 1\cdot 2 \cdot 3 \cdots n \cdot 1\cdot 2 \cdot 3 \cdots n} = \frac{(2n)!}{(n!)^2}$$
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