Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>\(\lim_{h \to 0} \left[\frac{1}{h\sqrt[3]{8+h}} - \frac{1}{2h}\right]\) is equal to</p>
<p>(a) \(\frac{1}{12}\)</p>
<p>(b) \(-\frac{1}{12}\)</p>
<p>(c) \(-\frac{16}{3}\)</p>
<p>(d) \(-\frac{1}{48}\)</p>
Step-by-Step Solution
Key Concept: Recognize this as the derivative of a cube root function at a specific point. Rewrite the expression as a limit definition: lim(h→0) [f(h)-f(0)]/h where f(x) = 1/∛x, then apply the derivative formula for x^(-1/3).
<p><strong>Step 1:</strong> Rewrite the limit by recognizing its structure:</p><p>lim(h→0) [1/(h∛(8+h)) - 1/(2h)] = lim(h→0) [(1/∛(8+h) - 1/2)/h]</p><p><strong>Step 2:</strong> This is the derivative definition of f(x) = 1/∛x at x = 8:</p><p>f(x) = x^(-1/3), so f'(x) = -1/3 · x^(-4/3)</p><p><strong>Step 3:</strong> Evaluate at x = 8:</p><p>f'(8) = -1/3 · (8)^(-4/3) = -1/3 · 1/(8^(4/3)) = -1/3 · 1/16 = -1/48</p><p>∴ Answer: D (which should be -1/48)</p>
Correct Answer: D