Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>If \(f(2) = 2\), \(f'(2) = 1\) then \(\lim_{x \to 2} \frac{2x^2 - 4f(x)}{x - 2} =\) ______</p>
Step-by-Step Solution
Key Concept: Recognize this limit contains f(x) and requires splitting into parts using the definition of derivative f'(2) = lim[x→2] (f(x)-f(2))/(x-2). Rearrange the numerator to isolate the f(x) term separately from the polynomial term.
<p><strong>Step 1:</strong> Rewrite the numerator by factoring and regrouping:</p><p>$$\lim_{x \to 2} \frac{2x^2 - 4f(x)}{x - 2} = \lim_{x \to 2} \frac{2x^2 - 8 - 4f(x) + 8}{x - 2}$$</p><p><strong>Step 2:</strong> Separate into two limits:</p><p>$$= \lim_{x \to 2} \frac{2(x^2 - 4)}{x - 2} + \lim_{x \to 2} \frac{8 - 4f(x)}{x - 2}$$</p><p><strong>Step 3:</strong> Evaluate the first limit (factor x² - 4):</p><p>$$= \lim_{x \to 2} \frac{2(x-2)(x+2)}{x - 2} + \lim_{x \to 2} \frac{4(2 - f(x))}{x - 2}$$</p><p>$$= \lim_{x \to 2} 2(x+2) + \lim_{x \to 2} \frac{-4(f(x) - 2)}{x - 2}$$</p><p><strong>Step 4:</strong> Apply values f(2)=2 and f'(2)=1 (recognizing the derivative definition):</p><p>$$= 2(4) - 4 \cdot \lim_{x \to 2} \frac{f(x) - f(2)}{x - 2}$$</p><p>$$= 8 - 4(1) = 4$$</p><p>∴ Answer: <strong>4</strong></p>
Correct Answer: 4