<p>If \(\lim_{x \to 0} \frac{1-\cos 2x \sin 5x}{x^2 \sin 3x} = \frac{10}{3}\)</p><p><em>State whether this statement is true or false.</em></p>
Step-by-Step Solution
Key Concept: Use Taylor expansions for small angles: cos(2x) ≈ 1 - 2x², sin(5x) ≈ 5x, sin(3x) ≈ 3x to simplify the limit and identify the correct form of the expression.
<p><strong>Step 1:</strong> Clarify the expression: The numerator is 1 - cos(2x)sin(5x), not (1 - cos(2x))sin(5x).</p><p><strong>Step 2:</strong> Apply Taylor expansions as x → 0:</p><p>• cos(2x) = 1 - 2x² + O(x⁴)</p><p>• sin(5x) = 5x - O(x³)</p><p>• sin(3x) = 3x - O(x³)</p><p><strong>Step 3:</strong> Calculate the numerator:</p><p>cos(2x)·sin(5x) = (1 - 2x² + ...)(5x + ...) = 5x - 10x³ + ...</p><p>Therefore: 1 - cos(2x)sin(5x) = 1 - 5x + 10x³ - ... ≠ polynomial starting with constant 1</p><p><strong>Step 4:</strong> Evaluate the limit:</p><p>lim(x→0) [1 - 5x + ...]/[3x²] = lim(x→0) 1/[3x²] - 5/(3x) + ...</p><p>This limit diverges to -∞ (not 10/3).</p><p><strong>Alternatively,</strong> if the expression were (1 - cos(2x))·sin(5x)/[x²sin(3x)]:</p><p>= [2x²]/[x²·3x] · 5x = (10/3) ✓</p><p>∴ <strong>The statement is FALSE.</strong> The given limit ≠ 10/3 for the expression as written.</p>
Correct Answer: A