X- 0 limsin 5x -sin 3x sin 6x -sin 4x

x→ 0 limsin 5x sin 3xsin 6x sin 4x = x→ 0 limsin 5x sin 3xxsin 6x sin 4xx = x→ 0lim 5(sin 5x5x) x→ 0lim 3(sin 3x3x)x→ 0lim 6(sin 6x6x) ...

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Antiderivative

x→ 0 limsin 5...

Question

$lim x \to 0 \frac{sin 5 x - sin 3 x}{sin 6 x - sin 4 x}$

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Solution

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Similar questions

\cos x + \cos 2 x + \cos 3 x = 0

\cos x + \cos 3 x - \cos 2 x = 0

\sin x + \sin 5 x = \sin 3 x

\cos x \cos 2 x \cos 3 x = \frac{1}{4}

\cos x + \sin x = \cos 2 x + \sin 2 x

\sin x + \sin 2 x + \sin 3 = 0

\sin x + \sin 2 x + \sin 3 x + \sin 4 x = 0

\sin 3 x - \sin x = 4 \cos^{2} x - 2

\sin 2 x - \sin 4 x + \sin 6 x = 0

sin 2 x - sin 4 x + sin 6 x = 0

sin 6 x + sin 4 x + sin 2 x = 0

x

y = \frac{cos x + cos 2 x + cos 3 x + cos 4 x + cos 5 x + cos 6 x + cos 7 x}{sin x + sin 2 x + sin 3 x + sin 4 x + sin 5 x + sin 6 x + sin 7 x},

y = e^{{({sin}^{2} x + {sin}^{4} x + {sin}^{6} x + \dots) {log}_{e} 2}}

x^{2} - 17 x + 16 = 0,

0 < x < \frac{π}{2} .

List I

List II .

\begin{matrix} List I & List II (a) & \frac{2 sin 2 x}{1 + {cos}^{2} x} & (p) & 1 (b) & \frac{2 sin x}{sin x + cos x} & (q) & \frac{4}{9} (c) & \infty \sum n = 1 (cot x)^{n} & (r) & \frac{2}{3} (d) & \infty \sum n = 1 n (cot x)^{2 n} & (s) & \frac{4}{3} \end{matrix}

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