lim_(x→0) ((a sin 3x − b sin 2x )/x^3 ) = (1/2) Find a and b .


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Question Number 162775 by john_santu last updated on 01/Jan/22

$\lim_{x \to 0} \frac{a \sin 3 x - b \sin 2 x}{x^{3}} = \frac{1}{2}$ $F i n d a a n d b .$
	Answered by mr W last updated on 01/Jan/22

	$= \lim_{x \to 0} \frac{a (3 x - \frac{27 x^{3}}{6} + o (x^{3})) - b (2 x - \frac{8 x^{3}}{6} + o (x^{3}))}{x^{3}}$ $= \lim_{x \to 0} \frac{(3 a - 2 b) x - \frac{(27 a - 8 b) x^{3}}{6} + o (x^{3})}{x^{3}} = \frac{1}{2}$ $\Rightarrow 3 a - 2 b = 0$ $\Rightarrow - \frac{27 a - 8 b}{6} = \frac{1}{2}$ $\Rightarrow a = - \frac{1}{5}, b = - \frac{3}{10}$
	Answered by bobhans last updated on 01/Jan/22
	$lim_(x→0) ((a sin 3x−3ax+3ax−2bx+2bx−b sin 2x)/x^3 ) = (1/2) lim_(x→0) ((a(sin 3x−3x)+x(3a−2b)+b(2x−sin 2x))/x^3 ) =(1/2) [ 3a=2b ] lim_(x→0) ((27a(sin 3x−3x))/((3x)^3 ))+lim_(x→0) ((8b(2x−sin 2x))/((2x)^3 )) = (1/2) −((27a)/6) +((8b)/6) = (1/2) ⇒8b−27a=3 ⇒ 4(3a)−27a = 3 → { ((−15a=3 ; a=−(1/5))),((b=−(3/(10)))) :}$
	$\lim_{x \to 0} \frac{a \sin 3 x - 3 ax + 3 ax - 2 bx + 2 bx - b \sin 2 x}{x^{3}} = \frac{1}{2}$ $\lim_{x \to 0} \frac{a (\sin 3 x - 3 x) + x (3 a - 2 b) + b (2 x - \sin 2 x)}{x^{3}} = \frac{1}{2}$ $[3 a = 2 b]$ $\lim_{x \to 0} \frac{27 a (\sin 3 x - 3 x)}{{(3 x)}^{3}} + \lim_{x \to 0} \frac{8 b (2 x - \sin 2 x)}{{(2 x)}^{3}} = \frac{1}{2}$ $- \frac{27 a}{6} + \frac{8 b}{6} = \frac{1}{2} \Rightarrow 8 b - 27 a = 3$ $\Rightarrow 4 (3 a) - 27 a = 3 \to {\begin{cases} - 15 a = 3; a = - \frac{1}{5} \\ b = - \frac{3}{10} \end{cases}$
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