<p>If
\[ \lim_{\alpha \to 0} \frac{e^{\cos(\alpha^n)-1}}{\alpha^m} = -\frac{1}{2} \]
then find the value of \(m - 2n\).</p>
Step-by-Step Solution
Key Concept: Use Taylor expansion of e^x and cos(x) around 0, then match powers of α to find the relationship between m and n that makes the limit finite and equal to -1/2.
<p><strong>Step 1:</strong> Expand cos(α^n) using Taylor series: cos(α^n) = 1 - (α^n)²/2 + (α^n)⁴/24 - ...</p><p><strong>Step 2:</strong> Therefore: cos(α^n) - 1 = -(α^(2n))/2 + (α^(4n))/24 - ...</p><p><strong>Step 3:</strong> Now expand e^(cos(α^n)-1) = e^(-(α^(2n))/2 + ...) using e^x = 1 + x + x²/2 + ...</p><p><strong>Step 4:</strong> e^(cos(α^n)-1) = 1 + [-(α^(2n))/2 + (α^(4n))/24 - ...] + [-(α^(2n))/2]²/2 + ...</p><p><strong>Step 5:</strong> e^(cos(α^n)-1) = 1 - (α^(2n))/2 + (α^(4n))/24 + (α^(4n))/8 + ...</p><p><strong>Step 6:</strong> The dominant term is -(α^(2n))/2, so: [e^(cos(α^n)-1) - 1]/α^m → [-(α^(2n))/2]/α^m = -(α^(2n-m))/2</p><p><strong>Step 7:</strong> For the limit to equal -1/2 (finite and non-zero), we need 2n - m = 0, so m = 2n</p><p><strong>Step 8:</strong> Therefore: m - 2n = 2n - 2n = <strong>0</strong></p>
Correct Answer: 0