Limits, Continuity & Differentiability
L'Hôpital's Rule and Limits
Grade 12
Question:
<p>Given that \(f(a) = g(a) = k\); \(f^n(a) \neq g^n(a)\) for some \(n \in \mathbb{N}\) and \[\lim_{x \to a} \frac{f(a)\cdot g(x) - f(a) - g(a)f(x) + g(a)}{g(x) - f(x)} = 4\] Find the value of \(k\).</p>
<p>1</p>
<p>2</p>
<p>3</p>
<p>4</p>
Step-by-Step Solution
Key Concept: Factor the numerator as f(a)[g(x)-g(a)] - g(a)[f(x)-f(a)] and recognize this as a linear combination of difference quotients converging to f(a)·g'(a) - g(a)·f'(a). Since f(a)=g(a)=k and the denominator g(x)-f(x)→0, use L'Hôpital's rule or analyze the leading terms of the Taylor expansions around x=a.
<p><strong>Step 1: Substitute f(a)=g(a)=k into the numerator</strong></p><p>Numerator = k·g(x) - k - k·f(x) + k = k[g(x) - f(x)]</p><p><strong>Step 2: Simplify the limit</strong></p><p>$$\lim_{x \to a} \frac{k[g(x) - f(x)]}{g(x) - f(x)} = \lim_{x \to a} k = k$$</p><p><strong>Step 3: Verify using Taylor expansion (alternative method)</strong></p><p>Let g(x) - g(a) = g'(a)(x-a) + o(x-a) and f(x) - f(a) = f'(a)(x-a) + o(x-a)</p><p>Numerator = k[g(x) - f(x)] = k[(g'(a) - f'(a))(x-a) + o(x-a)]</p><p>Denominator = [g'(a) - f'(a)](x-a) + o(x-a)</p><p>$$\lim_{x \to a} \frac{k(g'(a)-f'(a))(x-a)}{(g'(a)-f'(a))(x-a)} = k$$</p><p>(The condition f^n(a) ≠ g^n(a) ensures g'(a) ≠ f'(a), so the limit is well-defined)</p><p><strong>Step 4: Equate to given limit</strong></p><p>k = 4</p><p>∴ Answer: <strong>k = 4</strong> (Option D)</p>
Correct Answer: D