mathop {lim }limits_{x to 0} frac{1 - cos 6x} | vedclass

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(A) We know that $1 - \cos 	heta = 2 \sin^2(	heta/2)$.Therefore,$1 - \cos 6x = 2 \sin^2(3x)$.$\mathop {\lim }\limits_{x 	o 0} \frac{2 \sin^2(3x)}{x

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(A) We know that $1 - \cos 	heta = 2 \sin^2(	heta/2)$.Therefore,$1 - \cos 6x = 2 \sin^2(3x)$.$\mathop {\lim }\limits_{x 	o 0} \frac{2 \sin^2(3x)}{x} = \mathop {\lim }\limits_{x 	o 0} \left( 2 \cdot \frac{\sin^2(3x)}{x} ight)$.Multiply and divide by $9x$ to use the standard limit $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$:$= \mathop {\lim }\limits_{x 	o 0} \left( 2 \cdot \frac{\sin^2(3x)}{(3x)^2} \cdot 9x ight) = \mathop {\lim }\limits_{x 	o 0} \left( 2 \cdot 1^2 \cdot 9x ight) = 0$.

$\mathop {\lim }\limits_{x \to 0} \frac{1 - \cos 6x}{x} = $

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