mathop {lim }limits_{x to 0} left[ {frac{{sin | vedclass

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(C) Using the sum-to-product formula $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$,we have:$\sin(a+x) + \sin(a-x) = 2\sin a \cos x$.Substituting this into

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$-\sin a$

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(C) Using the sum-to-product formula $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$,we have:$\sin(a+x) + \sin(a-x) = 2\sin a \cos x$.Substituting this into the limit:$\mathop {\lim }\limits_{x 	o 0} \frac{2\sin a \cos x - 2\sin a}{x \sin x} = \mathop {\lim }\limits_{x 	o 0} \frac{2\sin a (\cos x - 1)}{x \sin x}$.$= -2\sin a \mathop {\lim }\limits_{x 	o 0} \left( \frac{1 - \cos x}{x^2} \cdot \frac{x}{\sin x} ight)$.Since $\mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$ and $\mathop {\lim }\limits_{x 	o 0} \frac{x}{\sin x} = 1$,we get:$= -2\sin a \cdot \frac{1}{2} \cdot 1 = -\sin a$.

$\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\sin (x + a) + \sin (a - x) - 2\sin a}}{{x\sin x}}} \right] = $

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