Limits, Continuity & Differentiability
Differentiability and Derivatives
Grade 12
Question:
<p>Let \(f(x)\) be a continuous and differentiable function such that \(\displaystyle\lim_{h\to 0}\frac{f(3+7h)-f(3+4h)}{h} = 4\). Then the value of \(f'(3)\) equals:</p>
<p>\(\dfrac{1}{3}\)</p>
<p>\(\dfrac{2}{3}\)</p>
<p>\(\dfrac{4}{3}\)</p>
<p>\(1\)</p>
Step-by-Step Solution
Key Concept: Use the definition of derivative and algebraic manipulation: rewrite the given limit by factoring out constants from the numerator to express it in terms of f'(3). The key is recognizing that both terms in the numerator involve f at points near 3, allowing us to separate them using limit properties.
<p><strong>Step 1:</strong> Start with the given limit and rewrite the numerator by adding and subtracting f(3):</p><p>$$\lim_{h\to 0}\frac{f(3+7h)-f(3+4h)}{h} = \lim_{h\to 0}\frac{[f(3+7h)-f(3)] - [f(3+4h)-f(3)]}{h}$$</p><p><strong>Step 2:</strong> Split the limit into two separate terms:</p><p>$$= \lim_{h\to 0}\frac{f(3+7h)-f(3)}{h} - \lim_{h\to 0}\frac{f(3+4h)-f(3)}{h}$$</p><p><strong>Step 3:</strong> Rewrite each term to match the derivative definition f'(3) = lim[f(3+u)-f(3)]/u as u→0:</p><p>$$= \lim_{h\to 0}\frac{f(3+7h)-f(3)}{7h} \cdot 7 - \lim_{h\to 0}\frac{f(3+4h)-f(3)}{4h} \cdot 4$$</p><p><strong>Step 4:</strong> Recognize that as h→0, both 7h→0 and 4h→0, so each term becomes f'(3):</p><p>$$= 7f'(3) - 4f'(3) = 3f'(3)$$</p><p><strong>Step 5:</strong> Given that this equals 4:</p><p>$$3f'(3) = 4$$</p><p>$$f'(3) = \frac{4}{3}$$</p><p>∴ Answer: C</p>
Correct Answer: C