Limits, Continuity & Differentiability
Limits Involving Stirling's Approximation
Grade 12
Question:
<p>The value of $\lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{\frac{3n^3 + 4}{4n^4 - 1}}$, $n \in \mathbb{N}$ is equal to:</p>
<p>(a) $\left(\frac{1}{e}\right)^{3/4}$</p>
<p>(b) $e^3$</p>
<p>(c) $e^{-1}$</p>
<p>(d) 0</p>
Step-by-Step Solution
Key Concept: Use Stirling's approximation for n! and logarithmic limit properties to convert the exponential limit into a manageable form. The exponent approaches a specific value as nāā, allowing us to evaluate the limit.
<p><strong>Step 1: Set up using logarithms.</strong> Let $L = \lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{\frac{3n^3 + 4}{4n^4 - 1}}$. Taking natural log:</p><p>$$\ln L = \lim_{n \to \infty} \frac{3n^3 + 4}{4n^4 - 1} \cdot \ln\left(\frac{n!}{n^n}\right)$$</p><p><strong>Step 2: Apply Stirling's approximation.</strong> By Stirling's formula: $n! \sim \sqrt{2\pi n} \cdot \left(\frac{n}{e}\right)^n$</p><p>Therefore: $$\ln(n!) \approx \frac{1}{2}\ln(2\pi n) + n\ln(n) - n$$</p><p>$$\ln\left(\frac{n!}{n^n}\right) = \ln(n!) - n\ln(n) \approx \frac{1}{2}\ln(2\pi n) - n$$</p><p><strong>Step 3: Simplify the exponent.</strong> For large n: $$\frac{3n^3 + 4}{4n^4 - 1} = \frac{3n^3}{4n^4}\left(1 + O(n^{-3})\right) = \frac{3}{4n}\left(1 + O(n^{-3})\right)$$</p><p><strong>Step 4: Combine and find the limit.</strong> $$\ln L = \lim_{n \to \infty} \frac{3}{4n} \cdot \left(\frac{1}{2}\ln(2\pi n) - n\right)$$</p><p>$$= \lim_{n \to \infty} \left[\frac{3\ln(2\pi n)}{8n} - \frac{3}{4}\right]$$</p><p>The first term goes to 0 as $n \to \infty$ (logarithm grows slower than linear). Therefore:</p><p>$$\ln L = 0 - \frac{3}{4} = -\frac{3}{4}$$</p><p><strong>Step 5: Exponentiate to find L.</strong> $$L = e^{-3/4} = \left(e^{-1}\right)^{3/4} = \left(\frac{1}{e}\right)^{3/4}$$</p><p><strong>ā“ Answer:</strong> a</p>
Correct Answer: a