<p>Let <em>A</em> =
\[
A = \lim_{x \to 0} \frac{2^{\tan x} - 2^{\sin x}}{x^3}
\]
Find the value of \(e^{4A}\).</p>
Step-by-Step Solution
Key Concept: Use Taylor series expansion of exponential functions: 2^u = e^(u·ln2) = 1 + u·ln2 + (u·ln2)²/2 + ... and expand tan(x) and sin(x) to sufficient order to find the leading term in the numerator.
<p><strong>Step 1:</strong> Use Taylor expansions: tan(x) = x + x³/3 + ... and sin(x) = x - x³/6 + ...</p><p><strong>Step 2:</strong> Compute tan(x) - sin(x) = x³/3 + x³/6 + ... = x³/2 + O(x⁵)</p><p><strong>Step 3:</strong> Use 2^u = e^(u·ln2) = 1 + u·ln2 + (u·ln2)²/2 + ..., so 2^(tan x) - 2^(sin x) = (tan x - sin x)·ln2 + O(x⁵) for the leading terms</p><p><strong>Step 4:</strong> Therefore, 2^(tan x) - 2^(sin x) = (x³/2)·ln2 + O(x⁵)</p><p><strong>Step 5:</strong> A = lim(x→0) [(x³/2)·ln2]/x³ = ln2/2</p><p><strong>Step 6:</strong> e^(4A) = e^(4·ln2/2) = e^(2ln2) = e^(ln4) = <strong>4</strong></p>
Correct Answer: 4