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Question Number 112497 by bemath last updated on 08/Sep/20

If lim_(x→0) ((sin 2x + asin x)/x^3 ) exist what is the value of a and the limit?

$$\mathrm{If}\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sin}\:\mathrm{2x}\:+\:\mathrm{asin}\:\mathrm{x}}{\mathrm{x}^{\mathrm{3}} }\:\mathrm{exist}\: \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}\:\mathrm{and}\:\mathrm{the}\:\mathrm{limit}? \\ $$

Answered by mathmax by abdo last updated on 08/Sep/20

let f(x) =((sin(2x)+asinx)/x^3 ) we have sinx ∼x−(x^3 /6) sin(2x)∼2x−(((2x)^3 )/6) =2x−(4/3)x^3 ⇒f(x) ∼ ((2x−(4/3)x^3 +ax−((ax^3 )/6))/x^3 ) =((2+a −x^2 ((4/3) +(a/6)))/x^2 ) so lim_(x→0) f(x) exist ⇔2+a =0 ⇔a=−2 and the limit is L =−((4/3)−(1/3)) =−1

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{sin}\left(\mathrm{2x}\right)+\mathrm{asinx}}{\mathrm{x}^{\mathrm{3}} }\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{sinx}\:\sim\mathrm{x}−\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{6}} \\ $$$$\mathrm{sin}\left(\mathrm{2x}\right)\sim\mathrm{2x}−\frac{\left(\mathrm{2x}\right)^{\mathrm{3}} }{\mathrm{6}}\:=\mathrm{2x}−\frac{\mathrm{4}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} \:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:\sim\:\frac{\mathrm{2x}−\frac{\mathrm{4}}{\mathrm{3}}\mathrm{x}^{\mathrm{3}} +\mathrm{ax}−\frac{\mathrm{ax}^{\mathrm{3}} }{\mathrm{6}}}{\mathrm{x}^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{2}+\mathrm{a}\:\:−\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{4}}{\mathrm{3}}\:+\frac{\mathrm{a}}{\mathrm{6}}\right)}{\mathrm{x}^{\mathrm{2}} }\:\:\mathrm{so}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \mathrm{f}\left(\mathrm{x}\right)\:\mathrm{exist}\:\Leftrightarrow\mathrm{2}+\mathrm{a}\:=\mathrm{0}\:\Leftrightarrow\mathrm{a}=−\mathrm{2} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{is}\:\mathrm{L}\:=−\left(\frac{\mathrm{4}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\right)\:=−\mathrm{1} \\ $$

Answered by bemath last updated on 08/Sep/20

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