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Evaluate $\lim \limits_{x \to 0} \bigg( \large\frac{\sin 2x+ \sin 6x}{\sin 5x -\sin 3x} \bigg)$

$\begin{array}{1 1}(A)\;4 \\(B)\;2 \\(C)\;3 \\(D)\;0 \end{array}$

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$\lim \limits_{x \to 0} \bigg( \large\frac{ \sin 2x + \sin 6x}{\sin 5x - \sin 3x} \bigg)$
$\qquad =\lim \limits_{x \to 0} \bigg( \large\frac{ 2 \sin 4x \cos 2x}{2 \cos 4x \sin x} \bigg)$
$\qquad =\lim \limits_{x \to 0} \bigg( \large\frac{ \sin 4x \cos 2x}{ \cos 4x \sin x} \bigg)$
$\qquad =\lim \limits_{x \to 0} \bigg \{ \large\frac{ \sin 4x}{4x} \times \frac{x}{\sin x }$$ \times \cos 2x \times \large\frac{1}{\cos 4x}$$ \times 4 \bigg \}$
$\qquad = 4 \times \lim \limits _{4x \to 0} \large\frac{\sin 4x}{4x} $$\times \lim \limits _{x \to 0} \bigg(\large\frac{x}{\sin x} \bigg) $$\times \lim \limits_{2x \to 0} \cos 2x \times \lim \limits _{4x \to 0} \large\frac{1}{\cos 4x}$
$\qquad= (4 \times 1 \times 1 \times 1 \times \large\frac{1}{1} )$
Hence A is the correct answer.
answered Apr 22, 2014 by meena.p
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