Limits, Continuity & Differentiability
Limits using L'Hôpital or differentiation
Grade 12
Question:
<p>Let \(f(2) = 4\) and \(f'(2) = 4\). Then \(\displaystyle\lim_{x \to 2} \frac{xf(2) - 2f(x)}{x - 2}\) is given by</p>
<p>2</p>
<p>\(-2\)</p>
<p>\(-4\)</p>
<p>3</p>
Step-by-Step Solution
Key Concept: Rewrite the limit to isolate f(x) and f(2) terms separately, then use the definition of derivative f'(2) = lim[h→0] (f(2+h) - f(2))/h. Substitute x = 2 + h to convert the given limit into a form involving f'(2).
<p><strong>Step 1:</strong> Rewrite the numerator by separating the terms involving f(2) and f(x):</p><p>$$\lim_{x \to 2} \frac{xf(2) - 2f(x)}{x - 2} = \lim_{x \to 2} \frac{xf(2) - 2f(2) + 2f(2) - 2f(x)}{x - 2}$$</p><p><strong>Step 2:</strong> Factor and split the limit:</p><p>$$= \lim_{x \to 2} \frac{(x-2)f(2) - 2[f(x) - f(2)]}{x - 2}$$</p><p>$$= \lim_{x \to 2} \left[f(2) - 2\cdot\frac{f(x) - f(2)}{x - 2}\right]$$</p><p><strong>Step 3:</strong> Apply the definition of derivative. As x → 2, the term $\frac{f(x) - f(2)}{x - 2} \to f'(2)$:</p><p>$$= f(2) - 2f'(2)$$</p><p><strong>Step 4:</strong> Substitute f(2) = 4 and f'(2) = 4:</p><p>$$= 4 - 2(4) = 4 - 8 = -4$$</p><p>∴ Answer: C</p>
Correct Answer: C