mathop {Limit}limits_{x to 0} {(cos 2x)^{3/x^ | vedclass

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(D) Let $L = \mathop {Limit}\limits_{x 	o 0} (\cos 2x)^{3/x^2}$.This is of the form $1^\infty$,so we use the formula $\mathop {Limit}\limits_{x 	o a

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$e^{-6}$

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(D) Let $L = \mathop {Limit}\limits_{x 	o 0} (\cos 2x)^{3/x^2}$.This is of the form $1^\infty$,so we use the formula $\mathop {Limit}\limits_{x 	o a} f(x)^{g(x)} = e^{\mathop {Limit}\limits_{x 	o a} g(x)(f(x)-1)}$.$L = e^{\mathop {Limit}\limits_{x 	o 0} \frac{3}{x^2}(\cos 2x - 1)}$.Using the identity $\cos 2x - 1 = -2\sin^2 x$,we get:$L = e^{\mathop {Limit}\limits_{x 	o 0} \frac{3}{x^2}(-2\sin^2 x)}$.$L = e^{-6 \mathop {Limit}\limits_{x 	o 0} (\frac{\sin x}{x})^2}$.Since $\mathop {Limit}\limits_{x 	o 0} \frac{\sin x}{x} = 1$,we have $L = e^{-6(1)^2} = e^{-6}$.

$\mathop {Limit}\limits_{x \to 0} {(\cos 2x)^{3/x^2}}$ has the value equal to . . . . . . .

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