Limits, Continuity & Differentiability
Limits using expansions
Grade 12
Question:
<p>The value of \(\displaystyle\lim_{x \to 0} \dfrac{\dfrac{x^2}{2} + 1 - \sqrt{1+x^2}}{\left(\cos x - e^{x^2}\right)\sin(x^2)}\) is equal to:</p>
<p>(a) \(\dfrac{1}{12}\)</p>
<p>(b) \(\dfrac{-1}{12}\)</p>
<p>(c) \(\dfrac{1}{6}\)</p>
<p>(d) \(\dfrac{-1}{6}\)</p>
Step-by-Step Solution
Key Concept: Use Taylor series expansions for √(1+x²), cos(x), and e^(x²) around x=0, then identify the leading order terms in numerator and denominator to evaluate the limit.
<p><strong>Step 1:</strong> Expand √(1+x²) using binomial series: √(1+x²) = 1 + x²/2 - x⁴/8 + O(x⁶)</p><p><strong>Step 2:</strong> Simplify numerator: x²/2 + 1 - √(1+x²) = x²/2 + 1 - (1 + x²/2 - x⁴/8 + ...) = x⁴/8 + O(x⁶)</p><p><strong>Step 3:</strong> Expand cos(x) and e^(x²): cos(x) = 1 - x²/2 + x⁴/24 + ..., and e^(x²) = 1 + x² + x⁴/2 + ...</p><p><strong>Step 4:</strong> Calculate cos(x) - e^(x²) = (1 - x²/2 + x⁴/24 + ...) - (1 + x² + x⁴/2 + ...) = -3x²/2 - 11x⁴/24 + O(x⁶)</p><p><strong>Step 5:</strong> Denominator: (cos(x) - e^(x²))sin(x²) = (-3x²/2 + O(x⁴)) · (x² + O(x⁶)) = -3x⁴/2 + O(x⁶)</p><p><strong>Step 6:</strong> Evaluate limit: lim = (x⁴/8)/(-3x⁴/2) = (1/8)/(-3/2) = (1/8) · (-2/3) = -1/12</p><p>∴ Answer: <strong>-1/12</strong></p>
Correct Answer: A