<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] By L'Hospital's rule, 
\begin{align*}
\lim_{x \to 1} \frac{1 - x^{n + k}}{1 - x^k} &= \lim_{x \to 1} \frac{-(n + k) x^{n + k}}{-k x^k} = \frac{n + k}{k}
\end{align*}
Hence the desired limit is
$$\frac{n + 1}{1} \cdot \frac{n + 2}{2} \cdot \frac{n + 3}{3} \cdots \frac{n + n}{n} = \frac{(2n)!}{(n!)^2}$$
Alternatively, factor out a common term of $1 - x$ to find
$$\frac{1 - x^{n + k}}{1 - x^k} = \frac{1 + x + x^2 + \dots + x^{n + k - 1}}{1 + x + x^2 + \dots + x^{k - 1}}$$
Now directly evaluate at $1$ to get $n + k$ in the numerator and $k$ in the denominator.
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