Limits, Continuity & Differentiability
Limits of Trigonometric Functions
Grade None
Question:
<p>Let \(\alpha\) and \(\beta\) be the distinct roots of \(ax^2 + bx + c = 0\), then \(\lim_{x \to \alpha} \dfrac{1 - \cos(ax^2 + bx + c)}{(x - \alpha)^2}\) is equal to</p>
<p>\(\dfrac{a^2}{2}(\alpha - \beta)^2\)</p>
<p>0</p>
<p>\(-\dfrac{a^2}{2}(\alpha - \beta)^2\)</p>
<p>\(\dfrac{1}{2}(\alpha - \beta)^2\)</p>
Step-by-Step Solution
Key Concept: Since α is a root of ax² + bx + c = 0, we have a(x - α)(x - β) as the factorization. Use the Taylor expansion: 1 - cos(u) ≈ u²/2 for small u, where u = ax² + bx + c = a(x - α)(x - β).
<p><strong>Step 1:</strong> Since α and β are roots of ax² + bx + c = 0, we can write:</p><p>ax² + bx + c = a(x - α)(x - β)</p><p><strong>Step 2:</strong> As x → α, let u = a(x - α)(x - β). Then u → 0 since (x - α) → 0.</p><p><strong>Step 3:</strong> Use the Taylor expansion: 1 - cos(u) = u²/2 + O(u⁴)</p><p>Therefore:</p><p>1 - cos(ax² + bx + c) = [a(x - α)(x - β)]²/2 + higher order terms</p><p><strong>Step 4:</strong> Substitute into the limit:</p><p>lim_{x → α} [a²(x - α)²(x - β)²/2] / (x - α)²</p><p><strong>Step 5:</strong> Cancel (x - α)²:</p><p>lim_{x → α} [a²(x - β)²/2]</p><p><strong>Step 6:</strong> Evaluate at x = α:</p><p>= a²(α - β)²/2</p><p>∴ Answer: <strong>a²(α - β)²/2</strong> (Option A)</p>
Correct Answer: A