Evaluating the unique limit

Question

How to evaluate this limit?

$$\lim_{x\to1}\frac{(1-x)(1-x^2)\cdots(1-x^{2n})}{((1-x)(1-x^2)\cdots(1-x^n))^2}$$

I tried l'Hospital's rule but couldn't proceed further.

Answer 1

$$ \begin{aligned} L &= \lim_{x\to1} \frac{(1-x)(1-x^2)\dots(1-x^{2n})}{((1-x)(1-x^2)\dots(1-x^n))^2} \\ &= \lim_{x\to1} \frac{(1-x^{n+1})(1-x^{n+2})\dots(1-x^{2n})}{(1-x)(1-x^2)\dots(1-x^n)} \\ &= \lim_{x\to1} \prod_{1\le k\le n}\frac{1-x^{n+k}}{1-x^k} \\ &= \prod_{1\le k\le n}\lim_{x\to1}\frac{1-x^{n+k}}{1-x^k} \\ &\qquad\qquad\qquad\textrm{now use l'Hospital or factorize $(1-x)$} \\ &= \prod_{1\le k\le n}\frac{n+k}{k} \\ &= \frac{(n+1)(n+2)\dots(2n)}{1\cdot 2\dots n} \\ &= \frac{(2n)!}{n!\cdot n!} \\ &=\binom {2n}n\ . \end{aligned} $$

Example using a computer algebra program in a special case, $n=10$:

var('k,x');
n = 10;
limit( prod( [ 1-x^k for k in [1..2*n] ] ) / prod( [1-x^k for k in [1..n] ] )^2, x=1 )

This produces...

sage: limit( prod( [ 1-x^k for k in [1..2*n] ] ) / prod( [1-x^k for k in [1..n] ] )^2, x=1 )
184756
sage: binomial(2*n,n)
184756

Answer 2

Guide:

One possible way:

Use $$1-x^{k+1}=(1-x)\sum_{i=0}^kx^i$$ to get rid of factor of $(1-x)$

Answer 3

Hint: You can cancel out factors in the denominator with factors in the numerator, leaving $$\lim_{x\to1}\frac{(1-x^{n+1})\cdots(1-x^{2n})}{(1-x)\cdots(1-x^n)}.$$

Now you can use the fact that $\lim_{x\to a}f(x)g(x)=(\lim_{x\to a}f(x))(\lim_{x\to a}g(x))$ assuming both limits on the right exist. This allows you to split up the inside part of the limit into 'easy' quotients that you can apply l'Hospital's rule to.

Answer 4

Note that

$$\frac{(1-x)(1-x^2)....(1-x^{2n})}{((1-x)(1-x^2)...(1-x^n))^2}\\ =\frac{(1-x)^{2n}}{(1-x)^{2n}}\frac{1\cdot (1+x)....(1+x+...+x^{2n-1})}{(1\cdot (1+x)...(1+x+...+x^{n-1}))^2}\to \frac{(2n)!}{(n!)^2}$$