Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>\(\lim_{h \to 0} \frac{2\left[\sqrt{3}\sin\left(\frac{\pi}{6}+h\right) - \cos\left(\frac{\pi}{6}+h\right)\right]}{\sqrt{3}h(\sqrt{3}\cos h - \sin h)}\) is equal to</p>
<p>(a) \(\frac{2}{3}\)</p>
<p>(b) \(\frac{4}{3}\)</p>
<p>(c) \(-2\sqrt{3}\)</p>
<p>(d) \(-\frac{4}{3}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the numerator equals 2sin(π/6 + h - π/6) = 2sin(h) after applying the sine difference formula, and the denominator contains √3h(√3cos h - sin h). Both numerator and denominator approach 0, requiring L'Hôpital's rule or algebraic simplification using sin h ≈ h as h → 0.
<p><strong>Step 1:</strong> Simplify the numerator using compound angle formula.</p><p>√3sin(π/6 + h) - cos(π/6 + h) = √3[sin(π/6)cos h + cos(π/6)sin h] - [cos(π/6)cos h - sin(π/6)sin h]</p><p>= √3[(1/2)cos h + (√3/2)sin h] - [(√3/2)cos h - (1/2)sin h]</p><p>= (√3/2)cos h + (3/2)sin h - (√3/2)cos h + (1/2)sin h = 2sin h</p><p><strong>Step 2:</strong> Substitute the simplified form into the limit.</p><p>lim<sub>h→0</sub> [2(2sin h)] / [√3h(√3cos h - sin h)] = lim<sub>h→0</sub> [4sin h] / [√3h(√3cos h - sin h)]</p><p><strong>Step 3:</strong> Evaluate the limit using sin h/h → 1 and direct substitution for the remaining part.</p><p>= [4 · 1] / [√3 · 1 · (√3 · 1 - 0)] = 4 / (√3 · √3) = 4/3</p><p>∴ Answer: <strong>D</strong></p>
Correct Answer: D