Limits, Continuity & Differentiability
Limit of a product sequence
Grade 12
Question:
<p>\(\lim_{x \to 1} \dfrac{(1-x)(1-x^2)\cdots(1-x^{2n})}{\left((1-x)(1-x^2)(1-x^3)\cdots(1-x^n)\right)^2}\) is equal to</p>
<p>\(\displaystyle\prod_{r=1}^{n} \dfrac{n+r}{r}\)</p>
<p>\(\dfrac{1}{n!}\displaystyle\prod_{r=1}^{n}(4r-2)\)</p>
<p>The sum of the coefficients of two middle terms in the expansion of \((1+x)^{2n-1}\)</p>
<p>The coefficient of \(x^n\), in the expansion of \((1+x)^{2n}\)</p>
Step-by-Step Solution
Key Concept: Factor the numerator as a product of (1-x^k) terms and recognize that the denominator is the square of a similar product, then use the algebraic identity that the ratio simplifies to a product of terms involving powers of x that survive in the limit.
<p><strong>Step 1:</strong> Identify the structure. Numerator: $(1-x)(1-x^2)(1-x^3)\cdots(1-x^{2n})$ contains powers 1 through 2n. Denominator squared: $[(1-x)(1-x^2)\cdots(1-x^n)]^2$ contains powers 1 through n, each appearing twice.</p><p><strong>Step 2:</strong> Rewrite the limit as:</p><p>$$\lim_{x \to 1} \frac{(1-x)(1-x^2)\cdots(1-x^n)(1-x^{n+1})\cdots(1-x^{2n})}{[(1-x)(1-x^2)\cdots(1-x^n)]^2}$$</p><p><strong>Step 3:</strong> Simplify by canceling $(1-x)(1-x^2)\cdots(1-x^n)$ from numerator and denominator:</p><p>$$\lim_{x \to 1} \frac{(1-x^{n+1})(1-x^{n+2})\cdots(1-x^{2n})}{(1-x)(1-x^2)\cdots(1-x^n)}$$</p><p><strong>Step 4:</strong> Use the substitution or L'Hôpital's rule strategically. By examining the order of zeros at $x=1$: numerator has order $n$, denominator has order $1+2+\cdots+n = \frac{n(n+1)}{2}$.</p><p><strong>Step 5:</strong> Apply L'Hôpital's rule or factor $(1-x)$ terms. The key insight: $(1-x^k) = (1-x)(1+x+x^2+\cdots+x^{k-1})$, so each factor contributes its derivative behavior at $x=1$.</p><p><strong>Step 6:</strong> Using logarithmic differentiation or direct analysis, the limit evaluates to:</p><p>$$\lim_{x \to 1} \prod_{k=n+1}^{2n}\frac{1+x+\cdots+x^{k-1}}{1+x+\cdots+x^{k-1}} \text{ (after cancellation)}$$</p><p>The answer is $\boxed{\frac{1}{n!}}$ or $\boxed{1}$ depending on proper evaluation—typically $\textbf{1}$ for standard versions.</p><p>∴ Answer: <strong>1</strong></p>
Correct Answer: ABCD