Tough limit to evaluate I am trying to solve this limit problem
$$\lim_{x\to 1} {(1-x)(1-x^2)....(1-x^{2n})\over[(1-x)(1-x^2)....(1-x^n)]^2}$$
I am not able to figure how to to convert it to a compact form. Any tips?
 A: $\require{cancel}$
Consider pairing terms together
$$\lim_{x\to 1}{(1-x)(1-x^{n+1})\over (1-x)^2}\cdot\lim_{x\to 1}{(1-x^2)(1-x^{n+2})\over (1-x^2)^2}\cdot\ldots\cdot\lim_{x\to 1}{(1-x^n)(1-x^{2n})\over (1-x^n)^2}$$
From here use the usual factorization
$$1-x^k=(1-x)(1+x+x^2+\ldots +x^{k-1})$$
Then each individual limit,
$$\lim_{x\to 1}{(1-x^k)(1-x^{n+k})\over (1-x^k)^2}=\lim_{x\to 1}{\cancel{(1+x+\ldots +x^{k-1})}(1+x+\ldots+x^{n+k-1})\over (1+x+\ldots + x^{k-1})^{\cancel{2}}}={n+k\over k}$$
This gives a total answer of
$$\prod_{k=1}^n{n+k\over k}={(n+1)(n+2)\ldots 2n\over (1)(2)\ldots (n-1)(n)}$$
by reversing the order we multiply them in this gives
$$\prod_{k=1}^n {2n-k\over k}={(2n)(2n-1)\ldots (n+1)\over n(n-1)\ldots (3)(2)(1)}$$
multiplying top and bottom by $n!$ gives
$${(2n!)\over (n!)^2}={2n\choose n}$$

Fun fact: This is called a $q$-binomial coefficient and is of great interest in algebraic combinatorics, so if you have any interest in the methods used to prove this, you may find the subject interesting, as it is certainly rich and interesting.
A: Outline: Note that $\frac{1-x^k}{1-x}=1+x+x^2+\cdots +x^{k-1}$. As $x\to 1$, this has limit $k$.
So when we divide top and bottom by $(1-x)^{2n}$ we get on top a product with limit $(2n)!$. At the bottom we get a product with limit $(n!)^2$. 
Thus the limit is the central binomial coefficient $\binom{2n}{n}$.
A: We have a factor of $1-x$ in every term on top and bottom and an equal number of terms. Therefore we can cancel. 
Using the property that 
$$(1-x^n)=(1-x)(1+x+x^2+\cdots+x^{n-1})$$
We can see now that
$$\lim_{x\to 1}\frac{(1-x^n)}{1-x}=n$$
Thus
$$\frac{(2n)!}{n!n!}=\binom{2n}{n}$$
A: The $r(1\le r\le n)$th term is $$T_r=\lim_{x\to1}\frac{(1-x^{2r-1})(1-x^{2r})}{(1-x^r)^2}$$
$$=\lim_{x\to1}\frac{x^{2r-1}-1}{x-1}\cdot\lim_{x\to1}\frac{x^{2r}-1}{x-1}\frac1{\left(\lim_{x\to1}\dfrac{x^r-1}{x-1}\right)^2}$$
Now for integer $\displaystyle a>-1,\lim_{x\to1}\dfrac{x^a-1}{x-1}=\dfrac{d(x^a)}{dx}_{(\text{at }x=1)}=\cdots=a$
or $\displaystyle \lim_{x\to1}\dfrac{x^a-1}{x-1}=\lim_{x\to1}(1+x+x^2+\cdots+x^{a-1})=a$
$$\implies T_r=\frac{(2r-1)2r}{r^2}$$
Hope you can take it home from here
