<p>Evaluate: \(\lim_{n \to \infty} \dfrac{(3n+3)(3n+2)(3n+1)(3n!)}{(n+1)^3 (3n)!}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator contains a product of three consecutive terms (3n+3)(3n+2)(3n+1) which can be rewritten as a factorial ratio. Specifically, (3n+3)! = (3n+3)(3n+2)(3n+1)(3n)!, allowing us to convert the expression into a form using binomial coefficients or factorial ratios that behave predictably as nāā.
<p><strong>Step 1:</strong> Rewrite the numerator by recognizing the factorial pattern.</p><p>(3n+3)(3n+2)(3n+1)(3n)! = (3n+3)!</p><p><strong>Step 2:</strong> Express the original limit as:</p><p>$$\lim_{n \to \infty} \frac{(3n+3)!}{(n+1)^3 (3n)!} = \lim_{n \to \infty} \frac{(3n+3)(3n+2)(3n+1)}{(n+1)^3}$$</p><p><strong>Step 3:</strong> Expand the numerator and compare degrees:</p><p>Numerator: $(3n+3)(3n+2)(3n+1) = 27n^3 + 54n^2 + 33n + 6$</p><p>Denominator: $(n+1)^3 = n^3 + 3n^2 + 3n + 1$</p><p><strong>Step 4:</strong> Divide both by the highest degree term $n^3$:</p><p>$$\lim_{n \to \infty} \frac{27n^3 + 54n^2 + 33n + 6}{n^3 + 3n^2 + 3n + 1} = \frac{27}{1} = 27$$</p><p>ā“ Answer: <strong>27</strong></p>
Correct Answer: 27