Limits, Continuity & Differentiability
Limits
Grade 12
Question:
<p>\(\lim_{x \to 0} \dfrac{\sin(\pi \cos^2 x)}{x^2}\) is equal to</p>
<p>\(-\pi\)</p>
<p>\(\pi\)</p>
<p>\(\dfrac{\pi}{2}\)</p>
<p>\(1\)</p>
Step-by-Step Solution
Key Concept: Use the Taylor expansion of cos²x near x=0 to simplify the argument of sine, then apply the standard limit sin(u)/u → 1 as u → 0.
<p><strong>Step 1:</strong> Expand cos²x using Taylor series near x=0:</p><p>cos²x = 1 - x²/2 + x⁴/12 + ...</p><p><strong>Step 2:</strong> Substitute into the limit:</p><p>sin(π cos²x) = sin(π(1 - x²/2 + x⁴/12 + ...))</p><p>= sin(π - πx²/2 + πx⁴/12 + ...)</p><p><strong>Step 3:</strong> Use sin(π - θ) = sin(θ):</p><p>sin(π - πx²/2 + ...) = sin(πx²/2 - πx⁴/12 + ...)</p><p><strong>Step 4:</strong> For small arguments, sin(u) ≈ u:</p><p>sin(πx²/2 - πx⁴/12 + ...) ≈ πx²/2 - πx⁴/12 + ...</p><p><strong>Step 5:</strong> Divide by x²:</p><p>lim(x→0) (πx²/2 - πx⁴/12 + ...)/x² = π/2</p><p>∴ Answer: <strong>B (π/2)</strong></p>
Correct Answer: B