mathop {lim }limits_{n to infty } cos left( { | vedclass

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(B) हम जानते हैं कि: $\cos A \cos 2A \cos 4A \dots \cos (2^{n-1}A) = \frac{\sin (2^n A)}{2^n \sin A}$.माना $A = \frac{x}{2^n}$. तब गुणनफल होगा:$\cos \

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$\frac{{\sin x}}{x}$

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(B) हम जानते हैं कि: $\cos A \cos 2A \cos 4A \dots \cos (2^{n-1}A) = \frac{\sin (2^n A)}{2^n \sin A}$.माना $A = \frac{x}{2^n}$. तब गुणनफल होगा:$\cos \left( \frac{x}{2^n} ight) \cos \left( \frac{x}{2^{n-1}} ight) \dots \cos \left( \frac{x}{2} ight) = \frac{\sin (2^n \cdot \frac{x}{2^n})}{2^n \sin (\frac{x}{2^n})} = \frac{\sin x}{2^n \sin (\frac{x}{2^n})}$.जब $n 	o \infty$ हो,तब सीमा लेने पर:$\mathop {\lim }\limits_{n 	o \infty } \frac{\sin x}{2^n \sin (\frac{x}{2^n})} = \sin x \cdot \mathop {\lim }\limits_{n 	o \infty } \frac{1}{2^n \sin (\frac{x}{2^n})}$.चूंकि $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,माना $	heta = \frac{x}{2^n}$. जैसे $n 	o \infty$,वैसे ही $	heta 	o 0$:$\mathop {\lim }\limits_{n 	o \infty } \frac{x/2^n}{\sin (x/2^n)} = 1$.अतः,व्यंजक $\frac{\sin x}{x} \cdot 1 = \frac{\sin x}{x}$ हो जाता है।

$\mathop {\lim }\limits_{n \to \infty } \cos \left( {\frac{x}{2}} \right)\cos \left( {\frac{x}{4}} \right)\cos \left( {\frac{x}{8}} \right) \dots \cos \left( {\frac{x}{{{2^n}}}} \right)$ का मान क्या है?

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