Question Number 196959 by cortano12 last updated on 05/Sep/23
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Answered by horsebrand11 last updated on 05/Sep/23
$$\:\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{sin}\:\mathrm{2x}−\mathrm{2x}}{\mathrm{x}^{\mathrm{3}} }\:\right)+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left(\frac{\mathrm{ax}+\mathrm{2x}}{\mathrm{x}^{\mathrm{3}} }\right)=\mathrm{1}−\mathrm{b} \\ $$$$\:=\:−\frac{\mathrm{8}}{\mathrm{6}}\:+\:\mathrm{0}\:=\:\mathrm{1}−\mathrm{b} \\ $$$$\:\Rightarrow\begin{cases}{\mathrm{a}=−\mathrm{2}}\\{\mathrm{b}=\frac{\mathrm{7}}{\mathrm{3}}}\end{cases} \\ $$
Answered by MM42 last updated on 05/Sep/23
$${lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{sin}\mathrm{2}{x}+{ax}+{bx}^{\mathrm{3}} }{{x}^{\mathrm{3}} }\: \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{2}{x}−\frac{\mathrm{4}}{\mathrm{3}}{x}^{\mathrm{3}} +{o}\left({x}\right)+{ax}+{bx}^{\mathrm{3}} }{{x}^{\mathrm{3}} }\: \\ $$$$={lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\left(\mathrm{2}+{a}\right){x}+\left({b}−\frac{\mathrm{4}}{\mathrm{3}}\right){x}^{\mathrm{3}} }{{x}^{\mathrm{3}} }\:=\mathrm{1} \\ $$$$\Rightarrow\begin{cases}{\mathrm{2}+{a}=\mathrm{0}\Rightarrow\:{a}=−\mathrm{2}}\\{{b}−\frac{\mathrm{4}}{\mathrm{3}}=\mathrm{1}\Rightarrow{b}=\frac{\mathrm{7}}{\mathrm{3}}}\end{cases} \\ $$$$ \\ $$