If $0 < x < π$ , then $\frac{sin 8 x + 7 sin 6 x + 18 sin 4 x + 12 sin 2 x}{sin 7 x + 6 sin 5 x + 12 sin 3 x}$ is equal to

6

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4

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3

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2

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8

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Solution

The correct option is C $2$
$lim x \to 0^{+} \frac{sin 8 x + 7 sin 6 x + 18 sin 4 x + 12 sin 2 x}{sin 7 x + 6 sin 5 x + 12 sin 3 x}$
$= lim x \to 0^{+} \frac{\frac{sin 8 x}{x} + 7 \frac{sin 6 x}{x} + 18 \frac{sin 4 x}{x} + 12 \frac{sin 2 x}{x}}{\frac{sin 7 x}{x} + 6 \frac{sin 5 x}{x} + 12 \frac{sin 3 x}{x}}$
$= \frac{8 + 7 \cdot 6 + 18 \cdot 4 + 12 \cdot 2}{7 + 6 \cdot 5 + 12 \cdot 3} = \frac{146}{73} = 2$

Looking at the options, only option D satisfies the limit condition.

Hence, Option D is correct

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