Limits, Continuity & Differentiability
Limits using L'Hôpital's Rule
Grade 12
Question:
<p>Find \(\displaystyle\lim_{x \to \alpha} \frac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}\), where \(x = \alpha\) is a root of \(ax^2 + bx + c = 0\).</p>
<p>\(0\)</p>
<p>\(\dfrac{a^2(\alpha - \beta)^2}{2}\)</p>
<p>\(\dfrac{(\alpha - \beta)^2}{2}\)</p>
<p>\(\dfrac{a^2(\alpha + \beta)^2}{2}\)</p>
Step-by-Step Solution
Key Concept: Since α is a root of ax² + bx + c = 0, the expression inside cosine vanishes at x = α. Use Taylor expansion of cos(u) ≈ 1 - u²/2 for small u, where u = ax² + bx + c, then factor this quadratic as a(x - α)(x - β).
<p><strong>Step 1:</strong> Since x = α is a root of ax² + bx + c = 0, we have ax² + bx + c = a(x - α)(x - β) where β is the other root.</p><p><strong>Step 2:</strong> Let u = ax² + bx + c = a(x - α)(x - β). As x → α, u → 0, so use Taylor expansion: cos(u) = 1 - u²/2 + O(u⁴)</p><p><strong>Step 3:</strong> Therefore, 1 - cos(u) = u²/2 + O(u⁴) = [a(x - α)(x - β)]²/2 + O(u⁴)</p><p><strong>Step 4:</strong> Substitute into the limit:</p><p>$$\lim_{x \to \alpha} \frac{a^2(x - \alpha)^2(x - \beta)^2/2 + O(u^4)}{(x - \alpha)^2}$$</p><p><strong>Step 5:</strong> Cancel (x - α)²:</p><p>$$\lim_{x \to \alpha} \frac{a^2(x - \beta)^2}{2} = \frac{a^2(\alpha - \beta)^2}{2}$$</p><p>∴ Answer: B (which equals <strong>a²(α - β)²/2</strong> or equivalently <strong>a²[(b² - 4ac)]/4a²</strong> = <strong>(b² - 4ac)/4</strong>)</p>
Correct Answer: B