Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>For continuity at <span>\(x = 0\)</span>, if <span>\(\lim_{x \to 0} \left[\dfrac{(e^x - 1)^2}{\sin\left(\dfrac{x}{k}\right) \cdot \ln\left(1 + \dfrac{x}{4}\right)}\right] = 12\)</span>, find <span>\(k\)</span>.</p>
<p>\(1\)</p>
<p>\(2\)</p>
<p>\(3\)</p>
<p>\(4\)</p>
Step-by-Step Solution
Key Concept: Use Taylor series expansions: e^x - 1 ≈ x + x²/2, sin(x/k) ≈ x/k, ln(1 + x/4) ≈ x/4 for small x, then match the limit to the given value.
<p><strong>Step 1:</strong> Expand using Taylor series for small x:</p><p>• e^x - 1 = x + x²/2 + x³/6 + ...</p><p>• (e^x - 1)² = (x + x²/2 + ...)² = x² + x³ + ... ≈ x² (for leading term)</p><p>• sin(x/k) = x/k - x³/(6k³) + ... ≈ x/k</p><p>• ln(1 + x/4) = x/4 - x²/32 + ... ≈ x/4</p><p><strong>Step 2:</strong> Substitute into the limit:</p><p>lim_{x→0} [(e^x - 1)²/(sin(x/k)·ln(1 + x/4))] = lim_{x→0} [x²/((x/k)·(x/4))]</p><p><strong>Step 3:</strong> Simplify the denominator:</p><p>= lim_{x→0} [x²/(x²/(4k))] = lim_{x→0} [x² · 4k/x²] = 4k</p><p><strong>Step 4:</strong> Set equal to the given limit:</p><p>4k = 12</p><p>k = 3</p><p>∴ Answer: C (k = 3)</p>
Correct Answer: C