<p>The value of <em>a</em> for which \(\lim_{x \to 0} \dfrac{\left(e^x - 1\right)^4}{\sin\!\left(\dfrac{x^2}{a^2}\right)\ln\!\left(1 + \dfrac{x^2}{2}\right)} = 8\) is</p>
Step-by-Step Solution
Key Concept: Use Taylor series expansions for each component: e^x - 1 ≈ x, sin(u) ≈ u, and ln(1+u) ≈ u for small arguments. The limit's form requires matching powers of x in numerator and denominator to find the coefficient.
<p><strong>Step 1:</strong> Expand each component using Taylor series as x → 0:</p><p>• (e^x - 1)⁴ ≈ x⁴</p><p>• sin(x²/a²) ≈ x²/a²</p><p>• ln(1 + x²/2) ≈ x²/2</p><p><strong>Step 2:</strong> Substitute into the limit:</p><p>$$\lim_{x \to 0} \frac{x^4}{\frac{x^2}{a^2} \cdot \frac{x^2}{2}} = \lim_{x \to 0} \frac{x^4}{\frac{x^4}{2a^2}}$$</p><p><strong>Step 3:</strong> Simplify:</p><p>$$= \lim_{x \to 0} \frac{x^4 \cdot 2a^2}{x^4} = 2a^2$$</p><p><strong>Step 4:</strong> Set equal to the given limit value:</p><p>$$2a^2 = 8$$</p><p>$$a^2 = 4$$</p><p>$$a = \pm 2$$</p><p><strong>∴ Answer: A (a = 2 or a = ±2, depending on options)</strong></p>
Correct Answer: A