Limits, Continuity & Differentiability
Limit of a sequence using logarithm
Grade 12
Question:
<p>The value of \(\lim_{n \to \infty} \left(\dfrac{n!}{n^n}\right)^{\frac{1}{n}}\) is equal to</p>
<p>\(1/e\)</p>
<p>\(e\)</p>
<p>\(e^2\)</p>
<p>\(\dfrac{1}{e^2}\)</p>
Step-by-Step Solution
Key Concept: Take logarithm of the expression and convert the limit into a Riemann sum form using Stirling's approximation or direct analysis of ln(n!). Recognize that (1/n)∑ln(k) from k=1 to n converges to ∫₀¹ ln(x)dx = -1, giving e^(-1) = 1/e.
<p><strong>Step 1:</strong> Let L = lim(n→∞) (n!/n^n)^(1/n). Take natural logarithm:</p><p>ln L = lim(n→∞) (1/n)·ln(n!/n^n) = lim(n→∞) (1/n)·[ln(n!) - n·ln(n)]</p><p><strong>Step 2:</strong> Rewrite as: ln L = lim(n→∞) (1/n)·∑(k=1 to n) ln(k) - ln(n)</p><p><strong>Step 3:</strong> Recognize (1/n)·∑(k=1 to n) ln(k) as a Riemann sum for ∫₀¹ ln(x)dx</p><p><strong>Step 4:</strong> Evaluate: ∫₀¹ ln(x)dx = [x·ln(x) - x]₀¹ = (0 - 1) - (0) = -1</p><p>Since ln(n) grows slower than n, the dominant term is -1</p><p><strong>Step 5:</strong> Therefore ln L = -1, so L = e^(-1) = 1/e</p><p>∴ Answer: <strong>1/e</strong></p>
Correct Answer: A