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Question Number 196959 by cortano12 last updated on 05/Sep/23

$\underset{―}{}$
	Answered by horsebrand11 last updated on 05/Sep/23
	$= lim_(x→0) (((sin 2x−2x)/x^3 ) )+lim_(x→0) (((ax+2x)/x^3 ))=1−b = −(8/6) + 0 = 1−b ⇒ { ((a=−2)),((b=(7/3))) :}$
	$= \lim_{x \to 0} (\frac{\sin 2 x - 2 x}{x^{3}}) + \lim_{x \to 0} (\frac{ax + 2 x}{x^{3}}) = 1 - b$ $= - \frac{8}{6} + 0 = 1 - b$ $\Rightarrow {\begin{cases} a = - 2 \\ b = \frac{7}{3} \end{cases}$
	Answered by MM42 last updated on 05/Sep/23
	$lim_(x→0) ((sin2x+ax+bx^3 )/x^3 ) =lim_(x→0) ((2x−(4/3)x^3 +o(x)+ax+bx^3 )/x^3 ) =lim_(x→0) (((2+a)x+(b−(4/3))x^3 )/x^3 ) =1 ⇒ { ((2+a=0⇒ a=−2)),((b−(4/3)=1⇒b=(7/3))) :}$
	${l i m}_{x \to 0} \frac{s i n 2 x + a x + {b x}^{3}}{x^{3}}$ $= {l i m}_{x \to 0} \frac{2 x - \frac{4}{3} x^{3} + o (x) + a x + {b x}^{3}}{x^{3}}$ $= {l i m}_{x \to 0} \frac{(2 + a) x + (b - \frac{4}{3}) x^{3}}{x^{3}} = 1$ $\Rightarrow {\begin{cases} 2 + a = 0 \Rightarrow a = - 2 \\ b - \frac{4}{3} = 1 \Rightarrow b = \frac{7}{3} \end{cases}$
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