<p>If \(\displaystyle\lim_{\alpha \to 0} \frac{e^{\cos(\alpha^n)} - e}{\alpha^m} = \frac{-e}{2}\) where \(m\) and \(n\) are positive integers greater than 1, then the value of \(\dfrac{m}{n}\) is:</p>
Step-by-Step Solution
Key Concept: Since cos(α^n) → 1 as α → 0, use the expansion e^(cos(α^n)) ≈ e·e^(cos(α^n)-1) and Taylor series: cos(α^n) - 1 ≈ -α^(2n)/2 to match the given limit value with the denominator power.
<p><strong>Step 1:</strong> As α → 0, we have cos(α^n) → cos(0) = 1, so the numerator approaches e^1 - e = 0. The limit is of the form 0/0.</p><p><strong>Step 2:</strong> Expand using Taylor series: e^(cos(α^n)) = e·e^(cos(α^n)-1)</p><p><strong>Step 3:</strong> Find cos(α^n) - 1: cos(α^n) = 1 - α^(2n)/2 + O(α^(4n))</p><p>Therefore: cos(α^n) - 1 = -α^(2n)/2 + O(α^(4n))</p><p><strong>Step 4:</strong> Expand e^(cos(α^n)-1): e^(-α^(2n)/2 + ...) = 1 - α^(2n)/2 + O(α^(4n))</p><p><strong>Step 5:</strong> Thus: e^(cos(α^n)) = e(1 - α^(2n)/2 + O(α^(4n))) = e - (e/2)α^(2n) + O(α^(4n))</p><p><strong>Step 6:</strong> The numerator becomes: e - (e/2)α^(2n) - e = -(e/2)α^(2n)</p><p><strong>Step 7:</strong> The limit is: lim(α→0) [-(e/2)α^(2n)]/α^m = -(e/2)·lim(α→0) α^(2n-m)</p><p><strong>Step 8:</strong> For this limit to equal -e/2, we need: 2n - m = 0, so m = 2n</p><p><strong>Step 9:</strong> Therefore: m/n = 2n/n = 2</p><p>∴ Answer: A</p>
Correct Answer: A