Limits, Continuity & Differentiability
Limits using Taylor/L'Hopital
Grade 12
Question:
<p><strong>155.</strong> 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 expansions for √(1+x²), cos(x), and e^(x²) around x=0, then apply L'Hôpital's rule or direct substitution after simplification to evaluate the indeterminate form.
<p><strong>Step 1: Expand numerator using Taylor series</strong></p><p>√(1+x²) = 1 + x²/2 - x⁴/8 + O(x⁶)</p><p>Numerator = x²/2 + 1 - (1 + x²/2 - x⁴/8 + ...) = x⁴/8 + O(x⁶)</p><p><strong>Step 2: Expand denominator using Taylor series</strong></p><p>cos(x) = 1 - x²/2 + x⁴/24 + ...</p><p>e^(x²) = 1 + x² + x⁴/2 + ...</p><p>cos(x) - e^(x²) = -3x²/2 - 11x⁴/24 + ...</p><p>sin(x²) = x² - x⁶/6 + ...</p><p><strong>Step 3: Multiply denominator factors</strong></p><p>(cos(x) - e^(x²))·sin(x²) = (-3x²/2 - 11x⁴/24 + ...)·(x² - x⁶/6 + ...)</p><p>= -3x⁴/2 + O(x⁶)</p><p><strong>Step 4: Compute the limit</strong></p><p>lim = (x⁴/8)/(-3x⁴/2) = (1/8)/(-3/2) = -1/12</p><p>∴ Answer: A</p>
Correct Answer: A