Limits, Continuity & Differentiability
Limit evaluation using series expansion
Grade 12
Question:
<p>If \(\lim_{x \to 0} \dfrac{\sqrt{1+x^2}\tan\sin\tan^{-1}x + 2\sqrt{1-x^2}\sin\tan\sin^{-1}x - 3x}{x^p} = L\) then choose the <strong>correct</strong> option</p>
<p>\(p = 3,\ L = \dfrac{-31}{60}\)</p>
<p>\(p = 5,\ L = \dfrac{-29}{60}\)</p>
<p>\(p = 5,\ L = \dfrac{-31}{60}\)</p>
<p>None of these</p>
Step-by-Step Solution
Key Concept: Expand the inverse trigonometric compositions using Taylor series near x=0: tan(sin(tan⁻¹x)) and sin(tan(sin⁻¹x)) have specific leading-order behaviors that must be matched with the denominator power p for a finite non-zero limit.
<p><strong>Step 1:</strong> Expand near x=0 using Taylor series:</p><p>• tan⁻¹x = x - x³/3 + O(x⁵)</p><p>• sin(tan⁻¹x) = x - x³/6 + O(x⁵)</p><p>• tan(sin⁻¹x) = x + x³/6 + O(x⁵)</p><p><strong>Step 2:</strong> Find tan(sin(tan⁻¹x)):</p><p>sin(tan⁻¹x) = x - x³/6 + ..., so tan(x - x³/6 + ...) = x - x³/6 + x³/3 + O(x⁵) = x + x³/6 + O(x⁵)</p><p><strong>Step 3:</strong> Find sin(tan(sin⁻¹x)):</p><p>tan(sin⁻¹x) = x + x³/6 + ..., so sin(x + x³/6 + ...) = x + x³/6 - x³/6 + O(x⁵) = x + O(x⁵)</p><p><strong>Step 4:</strong> Expand √(1+x²) = 1 + x²/2 - x⁴/8 + ... and √(1-x²) = 1 - x²/2 - x⁴/8 + ...</p><p><strong>Step 5:</strong> Numerator = (1 + x²/2 + ...)(x + x³/6 + ...) + 2(1 - x²/2 + ...)(x + O(x⁵)) - 3x</p><p>= x + x³/6 + x³/2 + 2x - x³ + O(x⁵) - 3x</p><p>= x + 2x - 3x + x³(1/6 + 1/2 - 1) + O(x⁵)</p><p>= 0 + x³(1/6 + 1/2 - 1) + O(x⁵) = x³(2/3 - 1) + O(x⁵) = -x³/3 + O(x⁵)</p><p><strong>Step 6:</strong> For finite non-zero limit: p = 3 and L = -1/3</p><p>∴ Answer: C</p>
Correct Answer: C