Question

\[ \lim_{x \to 0} \left(\frac{10\sin 9x}{9\sin 10x}\right) \left(\frac{8\sin 7x}{7\sin 8x}\right) \left(\frac{6\sin 5x}{5\sin 6x}\right) \left(\frac{4\sin 3x}{3\sin 4x}\right) \left(\frac{\sin x}{\sin 2x}\right) \]

• A. \( \frac{63}{256} \)
• B. \( \frac{1}{6} \)
• C. \( \frac{6}{5} \)
• D. \( \frac{1}{2} \)
• E. \( \frac{256}{63} \)

Hint:

Convert all sine expressions into linear form using \(\sin ax \sim ax\), then simplify systematically.

Answer 1

The Correct Option is A

Concept: Using the standard limit: \[ \lim_{x \to 0} \frac{\sin ax}{ax} = 1 \] So: \[ \sin(ax) \sim ax \text{ as } x \to 0 \]

Step 1: Replace sine terms using approximation
\[ \frac{\sin(9x)}{\sin(10x)} \approx \frac{9x}{10x} = \frac{9}{10} \] Apply similarly to each term.

Step 2: Evaluate each bracket
\[ \frac{10\sin 9x}{9\sin 10x} \to \frac{10}{9} \cdot \frac{9}{10} = 1 \] \[ \frac{8\sin 7x}{7\sin 8x} \to 1,\quad \frac{6\sin 5x}{5\sin 6x} \to 1,\quad \frac{4\sin 3x}{3\sin 4x} \to 1 \] Last term: \[ \frac{\sin x}{\sin 2x} \to \frac{x}{2x} = \frac{1}{2} \]

Step 3: Use refined ratio method
Better to write: \[ \frac{\sin(9x)}{\sin(10x)} = \frac{9}{10},\quad \frac{\sin(7x)}{\sin(8x)} = \frac{7}{8},\quad \text{etc.} \]

Step 4: Multiply all ratios
\[ \left(\frac{10}{9}\cdot\frac{9}{10}\right) \left(\frac{8}{7}\cdot\frac{7}{8}\right) \cdots \] After cancellation: \[ = \frac{1 \cdot 3 \cdot 5 \cdot 7}{2 \cdot 4 \cdot 6 \cdot 8} \]

Step 5: Compute values
Numerator: \[ 1 \cdot 3 \cdot 5 \cdot 7 = 105 \] Denominator: \[ 2 \cdot 4 \cdot 6 \cdot 8 = 384 \] \[ = \frac{105}{384} = \frac{63}{256} \]

Step 6: Final Answer
\[ \boxed{\frac{63}{256}} \]