Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>Let \(f(a) = g(a) = k\) and their \(n\)th derivatives \(f^n(a)\), \(g^n(a)\) exist and are not equal for some \(n\). Further, if \(\lim_{x \to a} \dfrac{f(a)g(x) - f(a) - g(a)f(x) + g(a)}{g(x) - f(x)} = 4\), then the value of \(k\) is</p>
<p>\(4\)</p>
<p>\(2\)</p>
<p>\(1\)</p>
<p>\(0\)</p>
Step-by-Step Solution
Key Concept: Apply L'Hôpital's rule repeatedly since f(a) = g(a) = k creates a 0/0 form, and identify the first derivative term that doesn't cancel due to the condition that f^n(a) ≠ g^n(a).
<p><strong>Step 1: Rewrite the numerator using f(a) = g(a) = k</strong></p><p>Numerator = f(a)g(x) - f(a) - g(a)f(x) + g(a) = f(a)[g(x) - 1] - g(a)[f(x) - 1]</p><p>Since f(a) = g(a) = k: = k[g(x) - 1] - k[f(x) - 1] = k[g(x) - f(x)]</p><p><strong>Step 2: Simplify the limit</strong></p><p>$$\lim_{x \to a} \frac{k[g(x) - f(x)]}{g(x) - f(x)} = k \lim_{x \to a} \frac{g(x) - f(x)}{g(x) - f(x)}$$</p><p>This only works if g(x) - f(x) remains in denominator. For 0/0 form, apply L'Hôpital's:</p><p><strong>Step 3: Apply L'Hôpital's rule</strong></p><p>$$\lim_{x \to a} \frac{k[g(x) - f(x)]}{g(x) - f(x)} = \lim_{x \to a} \frac{k[g'(x) - f'(x)]}{g'(x) - f'(x)} = k$$</p><p>But we're told f^n(a) ≠ g^n(a) for some n where all lower derivatives are equal. If f'(a) = g'(a), continue applying L'Hôpital's until derivatives differ.</p><p><strong>Step 4: Given that the limit equals 4</strong></p><p>From the structure: the limit reduces to k when the cancellation occurs at the first non-equal derivative level, giving k = 4.</p><p>∴ Answer: <strong>k = 4</strong> (Option B)</p>
Correct Answer: B