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Question Number 191887 by Rupesh123 last updated on 03/May/23

	Answered by mehdee42 last updated on 03/May/23

	$h o p \Rightarrow {l i m}_{x \to 0} \frac{s i n x c o s 2 x c o 3 x . . . c o s n x + 2 s i n 2 x c o s x c o s (3 x) . . . c o s (n x) + . . . + n s i n (n x) c o s x c o s (2 x) . . . c o s (n - 1) x}{2 x} =$ ${l i m}_{x \to 0} \frac{x (1 + 4 + 9 + . . . + n^{2})}{x} = \frac{n (n + 1) (2 n + 1)}{6} ✓$
	Commented by Subhi last updated on 03/May/23

	$i t m u s t b e 12 i n t h e d e n o m i n a t o r$
	Commented by mehdee42 last updated on 03/May/23

	$y e s . i s h o u l d h a v e p u t 2 x a t t h e b e g i n n i n g o f t h e d e n o m i n a t o r .$ $t h a n k y o u$
	Commented by Subhi last updated on 03/May/23

	$w e l c o m e$
	Answered by Subhi last updated on 03/May/23

	$a p p l y L {H o p i t a l s}^{^{'}} l a w$ ${l i m}_{x \to 0} \frac{s i n (x) . (c o s (2 x) . . . . . . . . . . c o s (n x) - c o s (x) . \frac{d}{d x} (c o s (2 x) . . . . . . . . c o s (n x))}{2 x}$ $w i t h s a m e w a y$ ${l i m}_{x \to 0} \frac{c o s (x) . c o s (2 x) . . . . . c o s (n x) + s i n (x) \frac{d}{d x} (c o s (2 x) . . . . c o s (n x)) + s i n (x) . \frac{d}{d x} (. . . . c o s (n x)) - c o s (x) . \frac{d^{2}}{{d x}^{2}} (c o s (2 x) . . . . c o s (n x))}{2}$ $w h e n s u b s t i t u t i n g w i t h x = 0$ $t e r m s i n c l u d i n g s i n (x) = 0$ $\frac{d^{2}}{{d x}^{2}} (c o s (2 x) . . . . c o s (n x))$ $\frac{d}{d x} = - 2 s i n (2 x) . . . . . c o s (n x) + c o s (2 x) . \frac{d}{d x} (c o s (3 x) . . . . c o s (n x))$ $\frac{d^{2}}{{d x}^{2}} = - 4 c o s (2 x) . . . . . c o s (n x) - 2 s i n (2 x) . \frac{d}{d x} (. . . . c o s (n x)) - 2 s i n (2 x) . \frac{d}{d x} (. . . . c o s (n x)) + c o s (2 x) . \frac{d^{2}}{{d x}^{2}} (c o s (3 x) . . . . c o s (n x))$ $- 4 + c o s (2 x) . \frac{d^{2}}{{d x}^{2}} (c o s (3 x) . . . c o s (n x))$ $b y t h e s a m e w a y$ ${l i m}_{x \to 0} \frac{1 - c o s (x) . . . . . . c o s (n x)}{x^{2}} = \frac{1 + 2^{2} + 3^{2} . . . . . . + n^{2}}{2}$ $= \frac{n (n + 1) (2 n + 1)}{2 \times 6}$
	Answered by qaz last updated on 04/May/23

	$\underset{x \to 0}{l i m} \frac{1 - \cos x \cdot \cos (2 x) \cdot . . . \cdot \cos (n x)}{x^{2}}$ $= - \underset{x \to 0}{l i m} \frac{\cos x \cdot \cos (2 x) \cdot . . . \cdot \cos (n x) - 1}{x^{2}}$ $= - \underset{x \to 0}{l i m} \frac{l n (1 + \cos x \cdot \cos (2 x) \cdot . . . \cdot \cos (n x) - 1)}{x^{2}}$ $= - \underset{x \to 0}{l i m} \frac{l n \cos x + l n \cos (2 x) + . . . + l n \cos (n x)}{x^{2}}$ $= - \underset{x \to 0}{l i m} \frac{- \frac{1}{2} x^{2} - \frac{1}{2} {(2 x)}^{2} - . . . - \frac{1}{2} {(n x)}^{2}}{x^{2}}$ $= \frac{n (n + 1) (2 n + 1)}{12}$
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