Question:

<p><strong>Step 1: Set up the problem</strong></p><p>We need $\lim_{x \to 0} \frac{(a-n)nx - \tan x}{\sin nx} = 0$ with $n \neq 0$.</p><p><strong>Step 2: Expand using Taylor series near x = 0</strong></p><p>Recall: $\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + ...$</p><p>And: $\sin nx = nx - \frac{(nx)^3}{6} + ... = nx - \frac{n^3x^3}{6} + ...$</p><p><strong>Step 3: Write the numerator</strong></p><p>Numerator = $(a-n)nx - \tan x$</p><p>$= (a-n)nx - \left(x + \frac{x^3}{3} + ...\right)$</p><p>$= [(a-n)n - 1]x - \frac{x^3}{3} + ...$</p><p><strong>Step 4: Write the denominator</strong></p><p>Denominator = $\sin nx = nx - \frac{n^3x^3}{6} + ...$</p><p><strong>Step 5: Apply the limit condition</strong></p><p>For the limit to equal 0 (not be indeterminate), the coefficient of x in the numerator must be zero:</p><p>$(a-n)n - 1 = 0$</p><p>$(a-n)n = 1$</p><p>$a - n = \frac{1}{n}$</p><p>$a = n + \frac{1}{n}$</p><p><strong>Step 6: Verify</strong></p><p>With $a = n + \frac{1}{n}$, the numerator becomes: $-\frac{x^3}{3} + ...$</p><p>The denominator is: $nx - \frac{n^3x^3}{6} + ...$</p><p>The limit is: $\lim_{x \to 0} \frac{-\frac{x^3}{3}}{nx} = \lim_{x \to 0} \frac{-x^2}{3n} = 0$ ✓</p><p><strong>∴ Answer: d</strong></p>