Limits, Continuity & Differentiability
Limits and Derivatives
Grade 12
Question:
<p>Let \(k = \lim_{x \to 0} \left( \frac{e^x(e^{nx}-1)}{e^x - 1} + e^x x^3 \right) = n\) and \(f(x) = e^x + e^{2x} + e^{3x} + \cdots + e^{nx} + e^x \cdot x^3\). If \(f'''(0) = 1^3 + 2^3 + 3^3 + \cdots + n^3 + 6 = 1302\), find the value of \(k + n\).</p>
Step-by-Step Solution
Key Concept: Recognize that the limit condition k = n creates a constraint on n, and the third derivative at x=0 extracts the sum of cubes formula. The sum 1³ + 2³ + ... + n³ = [n(n+1)/2]² allows you to solve for n directly.
<p><strong>Step 1:</strong> Evaluate the limit k = lim_{x→0} [e^x(e^(nx)-1)/(e^x-1) + e^x·x³]</p><p>Using series: e^x(e^(nx)-1)/(e^x-1) → (1+x+...)(nx+n²x²/2+...) / (x+x²/2+...) → n as x→0</p><p>The term e^x·x³ → 0 as x→0</p><p>Therefore: <strong>k = n</strong></p><p><strong>Step 2:</strong> Find f'''(0) from f(x) = e^x + e^(2x) + e^(3x) + ... + e^(nx) + e^x·x³</p><p>The third derivative of e^(jx) at x=0 is j³</p><p>The third derivative of e^x·x³ at x=0 is 6 (from the x³ term)</p><p>Therefore: f'''(0) = 1³ + 2³ + 3³ + ... + n³ + 6</p><p><strong>Step 3:</strong> Use the given condition f'''(0) = 1302</p><p>1³ + 2³ + 3³ + ... + n³ + 6 = 1302</p><p>1³ + 2³ + 3³ + ... + n³ = 1296</p><p><strong>Step 4:</strong> Apply the formula 1³ + 2³ + ... + n³ = [n(n+1)/2]²</p><p>[n(n+1)/2]² = 1296 = 36²</p><p>n(n+1)/2 = 36</p><p>n(n+1) = 72</p><p>n² + n - 72 = 0</p><p>(n+9)(n-8) = 0</p><p>Since n > 0: <strong>n = 8</strong></p><p><strong>Step 5:</strong> Calculate k + n</p><p>k = n = 8</p><p>∴ k + n = 8 + 8 = <strong>16</strong></p>
Correct Answer: 16