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Question: If $L = \lim_{x \to 1} \frac{(1-x)(1-x^2)(1-x^3)....(1-x^{2n})}{[(1-x)(1-x^2)(1-x^3)....(1-x^n)]^2}$...

If L=lim⁡x→1(1−x)(1−x2)(1−x3)....(1−x2n)[(1−x)(1−x2)(1−x3)....(1−xn)]2L = \lim_{x \to 1} \frac{(1-x)(1-x^2)(1-x^3)....(1-x^{2n})}{[(1-x)(1-x^2)(1-x^3)....(1-x^n)]^2} then L can be equal to

Answer

(2nn)\binom{2n}{n}

Explanation

Solution

The given limit is L=lim⁡x→1(1−x)(1−x2)(1−x3)....(1−x2n)[(1−x)(1−x2)(1−x3)....(1−xn)]2L = \lim_{x \to 1} \frac{(1-x)(1-x^2)(1-x^3)....(1-x^{2n})}{[(1-x)(1-x^2)(1-x^3)....(1-x^n)]^2}.

We can rewrite the term (1−xk)(1-x^k) using the factorization 1−xk=(1−x)(1+x+x2+...+xk−1)1-x^k = (1-x)(1+x+x^2+...+x^{k-1}).

Let the numerator be N(x)=∏k=12n(1−xk)N(x) = \prod_{k=1}^{2n} (1-x^k). Using the factorization, N(x)=∏k=12n[(1−x)(1+x+...+xk−1)]N(x) = \prod_{k=1}^{2n} [(1-x)(1+x+...+x^{k-1})]. N(x)=(1−x)2n∏k=12n(1+x+...+xk−1)N(x) = (1-x)^{2n} \prod_{k=1}^{2n} (1+x+...+x^{k-1}).

Let the denominator be D(x)=[∏k=1n(1−xk)]2D(x) = [\prod_{k=1}^{n} (1-x^k)]^2. Using the factorization, ∏k=1n(1−xk)=∏k=1n[(1−x)(1+x+...+xk−1)]\prod_{k=1}^{n} (1-x^k) = \prod_{k=1}^{n} [(1-x)(1+x+...+x^{k-1})]. ∏k=1n(1−xk)=(1−x)n∏k=1n(1+x+...+xk−1)\prod_{k=1}^{n} (1-x^k) = (1-x)^n \prod_{k=1}^{n} (1+x+...+x^{k-1}). So, D(x)=[(1−x)n∏k=1n(1+x+...+xk−1)]2=(1−x)2n[∏k=1n(1+x+...+xk−1)]2D(x) = [(1-x)^n \prod_{k=1}^{n} (1+x+...+x^{k-1})]^2 = (1-x)^{2n} [\prod_{k=1}^{n} (1+x+...+x^{k-1})]^2.

Now substitute these back into the expression for L: L=lim⁡x→1(1−x)2n∏k=12n(1+x+...+xk−1)(1−x)2n[∏k=1n(1+x+...+xk−1)]2L = \lim_{x \to 1} \frac{(1-x)^{2n} \prod_{k=1}^{2n} (1+x+...+x^{k-1})}{(1-x)^{2n} [\prod_{k=1}^{n} (1+x+...+x^{k-1})]^2}.

We can cancel the (1−x)2n(1-x)^{2n} terms, assuming x≠1x \ne 1: L=lim⁡x→1∏k=12n(1+x+...+xk−1)[∏k=1n(1+x+...+xk−1)]2L = \lim_{x \to 1} \frac{\prod_{k=1}^{2n} (1+x+...+x^{k-1})}{[\prod_{k=1}^{n} (1+x+...+x^{k-1})]^2}.

Now, we evaluate the limit of each term (1+x+...+xk−1)(1+x+...+x^{k-1}) as x→1x \to 1. lim⁡x→1(1+x+...+xk−1)=1+1+...+1\lim_{x \to 1} (1+x+...+x^{k-1}) = 1+1+...+1 (k times) =k= k.

Since the limit of each factor exists and is finite, the limit of the product is the product of the limits. The numerator limit is ∏k=12n(lim⁡x→1(1+x+...+xk−1))=∏k=12nk=1⋅2⋅3⋅...⋅(2n)=(2n)!\prod_{k=1}^{2n} (\lim_{x \to 1} (1+x+...+x^{k-1})) = \prod_{k=1}^{2n} k = 1 \cdot 2 \cdot 3 \cdot ... \cdot (2n) = (2n)!.

The denominator limit is [∏k=1n(lim⁡x→1(1+x+...+xk−1))]2=[∏k=1nk]2=[1⋅2⋅3⋅...⋅n]2=(n!)2[\prod_{k=1}^{n} (\lim_{x \to 1} (1+x+...+x^{k-1}))]^2 = [\prod_{k=1}^{n} k]^2 = [1 \cdot 2 \cdot 3 \cdot ... \cdot n]^2 = (n!)^2.

So, the limit L is: L=(2n)!(n!)2L = \frac{(2n)!}{(n!)^2}.

This expression is the definition of the binomial coefficient (2nn)\binom{2n}{n}. L=(2nn)L = \binom{2n}{n}.