Limits, Continuity & Differentiability
Limits using Taylor/expansion
Grade 12
Question:
<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>
<p>2</p>
<p>3</p>
<p>4</p>
<p>5</p>
Step-by-Step Solution
Key Concept: For the limit to exist and be finite with the given form, the numerator's Taylor expansion must have its leading term of order αᵐ. Since cos(αⁿ) ≈ 1 - αⁿ/2 for small α, we get e^(cos(αⁿ)) ≈ e·e^(-αⁿ/2) ≈ e(1 - αⁿ/2 + ...), making the numerator proportional to αⁿ. Thus m = n for a non-zero finite limit.
<p><strong>Step 1:</strong> Expand cos(αⁿ) near α = 0:</p><p>cos(αⁿ) = 1 - (αⁿ)²/2! + (αⁿ)⁴/4! - ... = 1 - αⁿ/2 + O(α⁴ⁿ)</p><p><strong>Step 2:</strong> Substitute into exponential:</p><p>e^(cos(αⁿ)) = e^(1 - α²ⁿ/2 + ...) = e · e^(-αⁿ/2 + O(α⁴ⁿ))</p><p><strong>Step 3:</strong> Expand e^(-αⁿ/2):</p><p>e^(-αⁿ/2) = 1 - αⁿ/2 + (αⁿ)²/8 - ... </p><p><strong>Step 4:</strong> Find numerator:</p><p>e^(cos(αⁿ)) - e = e[e^(-αⁿ/2) - 1] = e[-αⁿ/2 + (αⁿ)²/8 - ...] = -e·αⁿ/2 + O(α²ⁿ)</p><p><strong>Step 5:</strong> For finite non-zero limit:</p><p>lim(α→0) [e^(cos(αⁿ)) - e]/αᵐ = lim(α→0) [-e·αⁿ/2 + O(α²ⁿ)]/αᵐ</p><p>This equals -e/2 only when m = n</p><p><strong>Step 6:</strong> Calculate ratio:</p><p>m/n = n/n = 1</p><p>∴ Answer: A (m/n = 1)</p>
Correct Answer: A