If mathop {lim }limits_{n to infty } n cos le | vedclass

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(A) We are given the limit $L = \mathop {\lim }\limits_{n 	o \infty } n \cos \left( \frac{\pi }{4n} ight) \sin \left( \frac{\pi }{4n} ight)$.Mult

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$\frac{\pi }{4}$

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(A) We are given the limit $L = \mathop {\lim }\limits_{n 	o \infty } n \cos \left( \frac{\pi }{4n} ight) \sin \left( \frac{\pi }{4n} ight)$.Multiply and divide by $2$ to use the identity $\sin(2	heta) = 2 \sin 	heta \cos 	heta$:$L = \mathop {\lim }\limits_{n 	o \infty } \frac{n}{2} \left( 2 \sin \left( \frac{\pi }{4n} ight) \cos \left( \frac{\pi }{4n} ight) ight)$$L = \frac{1}{2} \mathop {\lim }\limits_{n 	o \infty } n \sin \left( \frac{\pi }{2n} ight)$Let $x = \frac{\pi }{2n}$. As $n 	o \infty$,$x 	o 0$. Also,$n = \frac{\pi }{2x}$.$L = \frac{1}{2} \mathop {\lim }\limits_{x 	o 0 } \left( \frac{\pi }{2x} ight) \sin(x)$$L = \frac{\pi }{4} \mathop {\lim }\limits_{x 	o 0 } \frac{\sin x}{x}$Since $\mathop {\lim }\limits_{x 	o 0 } \frac{\sin x}{x} = 1$,we get $L = \frac{\pi }{4} 	imes 1 = \frac{\pi }{4}$.Thus,$k = \frac{\pi }{4}$.

If $\mathop {\lim }\limits_{n \to \infty } n \cos \left( \frac{\pi }{4n} \right) \sin \left( \frac{\pi }{4n} \right) = k$,then $k$ is equal to

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