_ x 0 2x + 6x 5x - 3x = - Vidyadip

Question

$\mathop {\lim }\limits_{x \to 0} \frac{{\sin 2x + \sin 6x}}{{\sin 5x - \sin 3x}} = $

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Answer

d
(d) $\mathop {\lim }\limits_{x \to 0} \,\,\frac{{2\sin 4x\cos 2x}}{{2\sin x\cos 4x}} = \mathop {\lim }\limits_{x \to 0} 4\left( {\frac{{\sin 4x}}{{4x}}} \right)\,\left( {\frac{x}{{\sin x}}} \right)\frac{{\cos 2x}}{{\cos 4x}} = 4$.

Aliter : $\mathop {\lim }\limits_{x \to 0} \,\,\frac{{\frac{{2\,\,\sin 2x}}{{2x}} + \frac{{6\,\,\sin 6x}}{{6x}}}}{{\frac{{5\,\,\sin 5x}}{{5x}} - \frac{{3\,\,\sin 3x}}{{3x}}}} = \frac{{2 + 6}}{{5 - 3}} = 4.$

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