If lim _{x rightarrow 0} frac{cos 2x - cos 4x | vedclass

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(D) First,evaluate the limit for $k$: $\lim _{x \rightarrow 0} \frac{\cos 2x - \cos 4x}{1 - \cos 2x} = \lim _{x \rightarrow 0} \frac{(1 - 2\sin^2 x) -

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$\frac{1}{4}$

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(D) First,evaluate the limit for $k$: $\lim _{x ightarrow 0} \frac{\cos 2x - \cos 4x}{1 - \cos 2x} = \lim _{x ightarrow 0} \frac{(1 - 2\sin^2 x) - (1 - 8\sin^2 x \cos^2 x)}{2\sin^2 x}$ $= \lim _{x ightarrow 0} \frac{8\sin^2 x \cos^2 x - 2\sin^2 x}{2\sin^2 x} = \lim _{x ightarrow 0} (4\cos^2 x - 1) = 4(1)^2 - 1 = 3$. So,$k = 3$. Now,evaluate the second limit with $k = 3$: $\lim _{x ightarrow 3} \frac{x^3 - 27}{x^4 - 81} = \lim _{x ightarrow 3} \frac{(x - 3)(x^2 + 3x + 9)}{(x - 3)(x + 3)(x^2 + 9)} = \lim _{x ightarrow 3} \frac{x^2 + 3x + 9}{(x + 3)(x^2 + 9)} = \frac{9 + 9 + 9}{(3 + 3)(9 + 9)} = \frac{27}{6 	imes 18} = \frac{27}{108} = \frac{1}{4}$.

If $\lim _{x \rightarrow 0} \frac{\cos 2x - \cos 4x}{1 - \cos 2x} = k$,then $\lim _{x \rightarrow k} \frac{x^k - 27}{x^{k+1} - 81} = $

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