lim_(x→0) ((5sin x−sin 3x cos 2x−cos 3x sin 2x)/x^3 ) =?


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Question Number 162364 by cortano last updated on 29/Dec/21

$\lim_{x \to 0} \frac{5 \sin x - \sin 3 x \cos 2 x - \cos 3 x \sin 2 x}{x^{3}} = ?$
Commented by blackmamba last updated on 29/Dec/21

$L = \lim_{x \to 0} \frac{5 \sin x - \sin 5 x}{x^{3}}$ $L = \lim_{x \to 0} \frac{5 \sin x - \sin x + \sin x - \sin 5 x}{x^{3}}$ $L = \lim_{x \to 0} \frac{4 \sin x - 2 \cos 3 x \sin 2 x}{x^{3}}$ $L = \lim_{x \to 0} \frac{4 \sin x - 4 \sin x \cos x \cos 3 x}{x^{3}}$ $L = 4 \times \lim_{x \to 0} \frac{1 - \cos x \cos 3 x}{x^{2}}$ $L = 4 \times \lim_{x \to 0} \frac{1 - \cos x + \cos x - \cos x \cos 3 x}{x^{2}}$ $L = 4 \times \lim_{x \to 0} \frac{2 \sin^{2} (\frac{x}{2}) + \cos x (2 \sin^{2} (\frac{3 x}{2}))}{x^{2}}$ $L = 4 \times (\frac{1}{2} + \frac{9}{2}) = 20 .$
	Answered by aleks041103 last updated on 29/Dec/21

	$s i n 3 x c o s 2 x + c o s 3 x s i n 2 x = s i n 5 x$ $\Rightarrow a n s = \underset{x \to 0}{l i m} \frac{5 s i n x - s i n 5 x}{x^{3}}$ $u s e l^{'} h o p i t a l$ $a n s = \underset{x \to 0}{l i m} \frac{5 c o s x - 5 c o s 5 x}{3 x^{2}} = \frac{5}{3} \underset{x \to 0}{l i m} \frac{- s i n x + 5 s i n 5 x}{2 x} =$ $= \frac{5}{6} \underset{x \to 0}{l i m} \frac{- c o s x + 25 c o s 5 x}{1} = \frac{24.5}{6} = 20$ $A n s . 20$
	Answered by bobhans last updated on 29/Dec/21

	$\lim_{x \to 0} \frac{5 \sin x - \sin 5 x}{x^{3}} = \lim_{x \to 0} \frac{5 (x - \frac{x^{3}}{6}) - (5 x - \frac{{125 x}^{3}}{6})}{x^{3}}$ $= \lim_{x \to 0} \frac{\frac{120}{6} x^{3}}{x^{3}} = 20$
	Commented by aleks041103 last updated on 29/Dec/21

	$I t h i n k y o u s t i l l n e e d t o s h o w t h a t t h e$ $c o n t r i b u t i o n o f t h e r e s t o f t h e t e y l o r$ $s e r i e s i s 0$
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