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Question Number 165253 by LEKOUMA last updated on 28/Jan/22

	Answered by mindispower last updated on 28/Jan/22

	$s i n (x) = x - \frac{x^{3}}{6} + o (x^{3})$
	Answered by cortano1 last updated on 28/Jan/22

	$= - \frac{1}{18}$
	Answered by Ar Brandon last updated on 28/Jan/22
	$A=lim_(x→0) ((xsin(sinx)−sin^2 x)/x^6 ) =lim_(x→0) ((x(sinx−((sin^3 x)/6))−(x−(x^3 /6))^2 )/x^6 ) =lim_(x→0) ((x(x−(x^3 /6)−(1/6)(x−(x^3 /6))^3 )−(x^2 −(x^4 /3)+(x^6 /(36))))/x^6 ) =lim_(x→0) (1/x^6 ){x^2 −(x^4 /6)−(1/6)(x^4 −(x^6 /2)+(x^8 /(12))−(x^(10) /(216)))−x^2 +(x^4 /3)−(x^6 /(36))} =lim_(x→0) (1/x^6 )((x^6 /(18))−(x^8 /(72))+(x^(10) /(1296)))=lim_(x→0) ((1/(18))−(x^2 /(72))+(x^4 /(1296)))=(1/(18))$
	$A = \lim_{x \to 0} \frac{x \sin (\sin x) - \sin^{2} x}{x^{6}}$ $= \lim_{x \to 0} \frac{x (\sin x - \frac{\sin^{3} x}{6}) - {(x - \frac{x^{3}}{6})}^{2}}{x^{6}}$ $= \lim_{x \to 0} \frac{x (x - \frac{x^{3}}{6} - \frac{1}{6} {(x - \frac{x^{3}}{6})}^{3}) - (x^{2} - \frac{x^{4}}{3} + \frac{x^{6}}{36})}{x^{6}}$ $= \lim_{x \to 0} \frac{1}{x^{6}} {x^{2} - \frac{x^{4}}{6} - \frac{1}{6} (x^{4} - \frac{x^{6}}{2} + \frac{x^{8}}{12} - \frac{x^{10}}{216}) - x^{2} + \frac{x^{4}}{3} - \frac{x^{6}}{36}}$ $= \lim_{x \to 0} \frac{1}{x^{6}} (\frac{x^{6}}{18} - \frac{x^{8}}{72} + \frac{x^{10}}{1296}) = \lim_{x \to 0} (\frac{1}{18} - \frac{x^{2}}{72} + \frac{x^{4}}{1296}) = \frac{1}{18}$
	Commented by LEKOUMA last updated on 28/Jan/22

	$T h a n k$
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