mathop {lim }limits_{x to 0} frac{{sin 2x + s | vedclass

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(D) Using the sum-to-product formulas: $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \f

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Using the sum-to-product formulas: $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$ and $\sin A - \sin B = 2 \cos \frac{A+B}{2} \sin \frac{A-B}{2}$.$\mathop {\lim }\limits_{x 	o 0} \frac{{\sin 2x + \sin 6x}}{{\sin 5x - \sin 3x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{2 \sin 4x \cos (-2x)}}{{2 \cos 4x \sin x}} = \mathop {\lim }\limits_{x 	o 0} \frac{{\sin 4x \cos 2x}}{{\cos 4x \sin x}}$.Multiply and divide by $4x$ and $x$ to use the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$:$= \mathop {\lim }\limits_{x 	o 0} \left( \frac{\sin 4x}{4x} ight) \left( \frac{x}{\sin x} ight) \frac{4x \cos 2x}{x \cos 4x} = 1 	imes 1 	imes 4 	imes \frac{\cos(0)}{\cos(0)} = 4 	imes 1 = 4$.

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

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