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Question Number 38892 by gunawan last updated on 01/Jul/18

lim_(x→0) ((x(1+ a cos x)−b sin x)/x^5 )=1 find the value a and b

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\left(\mathrm{1}+\:{a}\:\mathrm{cos}\:{x}\right)−{b}\:\mathrm{sin}\:{x}}{{x}^{\mathrm{5}} }=\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:{a}\:\mathrm{and}\:{b} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jul/18

lim_(x→0) ((x+ax(1−(x^2 /(2!))+(x^4 /(4∙))−..)−b(x−(x^3 /(3!))+(x^5 /(5!))+..))/x^5 ) lim_(x→0) ((x(1+a−b)+a(((−x^3 )/(2!))+(x^5 /(4!))+..)−b(((−x^3 )/(3!))+(x^6 /(5!))+..))/x^5 ) lim_(x→0) ((1+a−b+a(−(x^2 /(2!))+(x^4 /(4!))+..)−b(((−x^2 )/(3!))+(x^5 /(5!))..))/x^4 ) 1+a−b=0 lim_(x→0) ((a(((−1)/(2!))+(x^2 /(4!))+..)−b(((−1)/(3!))+(x^3 /(5!))+..))/x^2 ) ((−a)/2)+(b/6)=0 b=3a 1+a−3a=0 a=(1/2) b=(3/2)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}+{ax}\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}\centerdot}−..\right)−{b}\left({x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}+..\right)}{{x}^{\mathrm{5}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{x}\left(\mathrm{1}+{a}−{b}\right)+{a}\left(\frac{−{x}^{\mathrm{3}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{4}!}+..\right)−{b}\left(\frac{−{x}^{\mathrm{3}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{6}} }{\mathrm{5}!}+..\right)}{{x}^{\mathrm{5}} } \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}+{a}−{b}+{a}\left(−\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{x}^{\mathrm{4}} }{\mathrm{4}!}+..\right)−{b}\left(\frac{−{x}^{\mathrm{2}} }{\mathrm{3}!}+\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}..\right)}{{x}^{\mathrm{4}} } \\ $$$$\mathrm{1}+{a}−{b}=\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{{a}\left(\frac{−\mathrm{1}}{\mathrm{2}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}!}+..\right)−{b}\left(\frac{−\mathrm{1}}{\mathrm{3}!}+\frac{{x}^{\mathrm{3}} }{\mathrm{5}!}+..\right)}{{x}^{\mathrm{2}} } \\ $$$$\frac{−{a}}{\mathrm{2}}+\frac{{b}}{\mathrm{6}}=\mathrm{0}\:\:{b}=\mathrm{3}{a}\:\:\: \\ $$$$\mathrm{1}+{a}−\mathrm{3}{a}=\mathrm{0}\:\:{a}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:{b}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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