<p>Evaluate \(\lim_{x \to \infty} \dfrac{x^a \sin\dfrac{1}{x^a} + x^2}{1 + |x|^3}\) when \(x \to \infty\).</p>
Step-by-Step Solution
Key Concept: For large x, decompose the numerator using the Taylor expansion sin(1/x^a) ≈ 1/x^a - 1/(6x^3a) + ... and compare the highest degree terms in numerator and denominator as x → ∞.
<p><strong>Step 1:</strong> For x → ∞, use the Taylor expansion: sin(1/x^a) = 1/x^a - 1/(6x^3a) + O(1/x^5a)</p><p><strong>Step 2:</strong> Therefore: x^a sin(1/x^a) = x^a · [1/x^a - 1/(6x^3a) + ...] = 1 - 1/(6x^2a) + ...</p><p><strong>Step 3:</strong> The numerator becomes: 1 - 1/(6x^2a) + x^2 = x^2 + 1 - 1/(6x^2a) + O(1/x^4a)</p><p><strong>Step 4:</strong> For x → +∞, we have |x|^3 = x^3, so the denominator is 1 + x^3</p><p><strong>Step 5:</strong> The limit is: lim(x→∞) (x^2 + 1 + ...)/(x^3 + 1) where the highest degree term in denominator dominates</p><p><strong>Step 6:</strong> Dividing by x^3: lim(x→∞) (1/x + 1/x^3 + ...)/(1 + 1/x^3) = 0/1 = 0</p><p>∴ <strong>Answer: 0</strong></p>
Correct Answer: 0