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Question-196959




Question Number 196959 by cortano12 last updated on 05/Sep/23
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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))) :}
$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\mathrm{2x}−\mathrm{2x}}{\mathrm{x}^{\mathrm{3}} }\:\right)+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{ax}+\mathrm{2x}}{\mathrm{x}^{\mathrm{3}} }\right)=\mathrm{1}−\mathrm{b} \\ $$$$\:=\:−\frac{\mathrm{8}}{\mathrm{6}}\:+\:\mathrm{0}\:=\:\mathrm{1}−\mathrm{b} \\ $$$$\:\Rightarrow\begin{cases}{\mathrm{a}=−\mathrm{2}}\\{\mathrm{b}=\frac{\mathrm{7}}{\mathrm{3}}}\end{cases} \\ $$
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))) :}
$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{sin}\mathrm{2}{x}+{ax}+{bx}^{\mathrm{3}} }{{x}^{\mathrm{3}} }\: \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}{x}−\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{3}} +{o}\left({x}\right)+{ax}+{bx}^{\mathrm{3}} }{{x}^{\mathrm{3}} }\: \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\left(\mathrm{2}+{a}\right){x}+\left({b}−\frac{\mathrm{4}}{\mathrm{3}}\right){x}^{\mathrm{3}} }{{x}^{\mathrm{3}} }\:=\mathrm{1} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2}+{a}=\mathrm{0}\Rightarrow\:{a}=−\mathrm{2}}\\{{b}−\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{1}\Rightarrow{b}=\frac{\mathrm{7}}{\mathrm{3}}}\end{cases} \\ $$$$ \\ $$

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