<p>If \(\lim_{x \to 0} \frac{x(1 + m\cos x) - n\sin x}{x^3} = 1\) then \(m =\) ______, \(n =\) ______</p>
Step-by-Step Solution
Key Concept: For the limit to exist and equal 1 as x→0, the numerator must have a zero of order exactly 3 at x=0. Use Taylor series expansions of cos x and sin x to match coefficients of powers of x.
<p><strong>Step 1:</strong> Expand using Taylor series: cos x = 1 - x²/2 + x⁴/24 - ... and sin x = x - x³/6 + ...</p><p><strong>Step 2:</strong> Substitute into numerator:<br>x(1 + m cos x) - n sin x = x(1 + m(1 - x²/2 + ...)) - n(x - x³/6 + ...)<br>= x + mx - mx³/2 - nx + nx³/6 + ...<br>= x(1 + m - n) + x³(-m/2 + n/6) + ...</p><p><strong>Step 3:</strong> For the limit to exist finitely, the coefficient of x must be zero:<br>1 + m - n = 0 ⟹ n = 1 + m</p><p><strong>Step 4:</strong> The limit becomes:<br>lim(x→0) [x³(-m/2 + n/6)]/x³ = -m/2 + n/6 = 1</p><p><strong>Step 5:</strong> Solve the system:<br>From step 3: n = 1 + m<br>Substitute into step 4: -m/2 + (1 + m)/6 = 1<br>-m/2 + 1/6 + m/6 = 1<br>-3m/6 + m/6 = 5/6<br>-2m/6 = 5/6<br>m = -5/2</p><p><strong>Step 6:</strong> n = 1 + m = 1 - 5/2 = -3/2</p><p>∴ <strong>m = -5/2, n = -3/2</strong></p>
Correct Answer: m = -5/2, n = -3/2