Evaluate: $\lim\limits_{x \to 0} \; \dfrac{\sin 2x + \sin 6x}{\sin 5x - \sin 3x}$
$\sin C + \sin D = 2 \sin \left(\dfrac{C+D}{2}\right) \cos \left(\dfrac{C-D}{2}\right)$
$\sin C - \sin D = 2 \cos \left(\dfrac{C+D}{2}\right) \sin \left(\dfrac{C-D}{2}\right)$
$\begin{aligned}
\therefore \lim\limits_{x \to 0} \;\; \dfrac{\sin 2x + \sin 6x}{\sin 5x - \sin 3x} & = \lim\limits_{x \to 0} \dfrac{2 \sin \left(\dfrac{6x+2x}{2}\right) \cos \left(\dfrac{6x-2x}{2}\right)}{2 \cos \left(\dfrac{5x+3x}{2} \right) \sin \left(\dfrac{5x-3x}{2}\right)} \\
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& = \lim\limits_{x \to 0} \; \dfrac{\sin 4x \cos 2x}{\cos 4x \sin x} \\
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& = \dfrac{\lim\limits_{4x \to 0}\; \dfrac{\sin 4x}{4x}\times 4 \times \lim\limits_{x \to 0} \; \cos 2x}{\lim\limits_{x \to 0} \; \cos 4x \times \lim\limits_{x \to 0} \; \dfrac{\sin x}{x}} \\
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& \left[\text{Note: Dividing numerator and denominator by x} \right]\\
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& \left[\text{Note: As } x \to 0, 4x \to 0\right] \\
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& = \dfrac{1 \times 4 \times 1}{1 \times 1} = 4
\end{aligned}$