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lim-x-0-a-sin-3x-b-sin-2x-x-3-1-2-Find-a-and-b-




Question Number 162775 by john_santu last updated on 01/Jan/22
   lim_(x→0)  ((a sin 3x − b sin 2x )/x^3 ) = (1/2)   Find a and b .
$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{a}\:\mathrm{sin}\:\mathrm{3}{x}\:−\:{b}\:\mathrm{sin}\:\mathrm{2}{x}\:}{{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:{Find}\:{a}\:{and}\:{b}\:. \\ $$
Answered by mr W last updated on 01/Jan/22
=lim_(x→0) ((a(3x−((27x^3 )/6)+o(x^3 ))−b(2x−((8x^3 )/6)+o(x^3 )))/x^3 )  =lim_(x→0) (((3a−2b)x−(((27a−8b)x^3 )/6)+o(x^3 ))/x^3 )=(1/2)  ⇒3a−2b=0  ⇒−((27a−8b)/6)=(1/2)  ⇒a=−(1/5), b=−(3/(10))
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}\left(\mathrm{3}{x}−\frac{\mathrm{27}{x}^{\mathrm{3}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{3}} \right)\right)−{b}\left(\mathrm{2}{x}−\frac{\mathrm{8}{x}^{\mathrm{3}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{3}} \right)\right)}{{x}^{\mathrm{3}} } \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{3}{a}−\mathrm{2}{b}\right){x}−\frac{\left(\mathrm{27}{a}−\mathrm{8}{b}\right){x}^{\mathrm{3}} }{\mathrm{6}}+{o}\left({x}^{\mathrm{3}} \right)}{{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{a}−\mathrm{2}{b}=\mathrm{0} \\ $$$$\Rightarrow−\frac{\mathrm{27}{a}−\mathrm{8}{b}}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{5}},\:{b}=−\frac{\mathrm{3}}{\mathrm{10}} \\ $$
Answered by bobhans last updated on 01/Jan/22
  lim_(x→0)  ((a sin 3x−3ax+3ax−2bx+2bx−b sin 2x)/x^3 ) = (1/2)   lim_(x→0)  ((a(sin 3x−3x)+x(3a−2b)+b(2x−sin 2x))/x^3 ) =(1/2)   [ 3a=2b ]    lim_(x→0)  ((27a(sin 3x−3x))/((3x)^3 ))+lim_(x→0)  ((8b(2x−sin 2x))/((2x)^3 )) = (1/2)   −((27a)/6) +((8b)/6) = (1/2) ⇒8b−27a=3  ⇒ 4(3a)−27a = 3 → { ((−15a=3 ; a=−(1/5))),((b=−(3/(10)))) :}
$$\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{a}\:\mathrm{sin}\:\mathrm{3x}−\mathrm{3ax}+\mathrm{3ax}−\mathrm{2bx}+\mathrm{2bx}−\mathrm{b}\:\mathrm{sin}\:\mathrm{2x}}{\mathrm{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{a}\left(\mathrm{sin}\:\mathrm{3x}−\mathrm{3x}\right)+\mathrm{x}\left(\mathrm{3a}−\mathrm{2b}\right)+\mathrm{b}\left(\mathrm{2x}−\mathrm{sin}\:\mathrm{2x}\right)}{\mathrm{x}^{\mathrm{3}} }\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\left[\:\mathrm{3a}=\mathrm{2b}\:\right]\: \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{27a}\left(\mathrm{sin}\:\mathrm{3x}−\mathrm{3x}\right)}{\left(\mathrm{3x}\right)^{\mathrm{3}} }+\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{8b}\left(\mathrm{2x}−\mathrm{sin}\:\mathrm{2x}\right)}{\left(\mathrm{2x}\right)^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:−\frac{\mathrm{27a}}{\mathrm{6}}\:+\frac{\mathrm{8b}}{\mathrm{6}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{8b}−\mathrm{27a}=\mathrm{3} \\ $$$$\Rightarrow\:\mathrm{4}\left(\mathrm{3a}\right)−\mathrm{27a}\:=\:\mathrm{3}\:\rightarrow\begin{cases}{−\mathrm{15a}=\mathrm{3}\:;\:\mathrm{a}=−\frac{\mathrm{1}}{\mathrm{5}}}\\{\mathrm{b}=−\frac{\mathrm{3}}{\mathrm{10}}}\end{cases} \\ $$

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