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Question Number 105410 by john santu last updated on 28/Jul/20

$$\underset{x\to 0}{\lim }\frac{x(1+a\cos x)-b\sin x}{x^5}=1$$ $$finda\mathrm{\&}b$$

Answered by bobhans last updated on 29/Jul/20

$$\underset{x\to 0}{\lim }\frac{x(1+a(1-\frac{1}{2}{x}^2+\frac{1}{24}{x}^4))-b(x-\frac{1}{6}{x}^3+\frac{x^5}{120})}{x^5}=1$$ $$\underset{x\to 0}{\lim }\frac{x(1+a-\frac{1}{2}{ax}^2+\frac{1}{24}{ax}^4)-(bx-\frac{1}{6}{bx}^3+\frac{b}{120}{x}^5)}{x^5}=1$$ $$\underset{x\to 0}{\lim }\frac{(1+a-b)x+(\frac{1}{6}b-\frac{1}{2}a)x^3+(\frac{1}{24}a-\frac{1}{120}b)x^5}{x^5}=1$$ $$\left(1\right)1+a-b=0;a-b=-1$$ $$\left(2\right)\frac{1}{6}b-\frac{1}{2}a=0;b=3a$$ $$\left(3\right)\frac{1}{24}a-\frac{1}{120}b=1;a=\frac{1}{5}b$$ $$thequestioninconsistent$$

Commented by1549442205PVT last updated on 29/Jul/20

$$\mathrm{I}\mathrm{think}\mathrm{that}\mathrm{in}\mathrm{the}\mathrm{question}\mathrm{need}\mathrm{to}\mathrm{give}$$ $$\mathrm{three}\mathrm{unknown}$$

Answered by Dwaipayan Shikari last updated on 29/Jul/20

$$\underset{x\to 0}{\lim }\frac{x(1+acosx)-bsinx}{x^5}=\frac{xa(-sinx)+acosx-bcosx}{5x^4}$$ $$=\frac{xa(-cosx)+a(-sinx)+a(-sinx)+bsinx}{20x^3}$$ $$=\frac{xasinx+a(-cosx)+a(-cosx)+a(-cosx)+bcosx}{60x^2}$$ $$=\frac{xacosx+asinx+3asinx-bsinx}{120x}or\frac{xa(-sinx)+acosx+4acosx-bcosx}{120}$$ $$=\frac{acosx}{120}+\frac{ax}{120x}+\frac{3ax}{120x}-\frac{bx}{120x}or\frac{5acosx-bcosx}{120}$$ $$=\frac{a}{24}-\frac{b}{120}$$ $$\Rightarrow \frac{a}{24}-\frac{b}{120}=1$$ $$5a-b=120\to \left(1\right)$$ $$\underset{x\to 0}{\lim }\frac{x(1+acosx)-bsinx}{x^5}=1$$ $$1+acosx-b=0$$ $$1+a=b\to \left(2\right)$$ $$5a-1-a=120$$ $$a=\frac{121}{4}$$ $$b=\frac{125}{4}$$ $$$$

Commented by1549442205PVT last updated on 30/Jul/20

$$\mathrm{♮}\lim \frac{x(1+acosx)-bsinx}{x^5}=1$$ $$1+acosx-b=0\epsilon \left(1\right)$$ $$\mathrm{I}\mathrm{think}\mathrm{that}\mathrm{ocurred}\mathrm{a}\mathrm{mistake}\mathrm{at}\mathrm{this}$$ $$\mathrm{place}.\mathrm{It}\mathrm{is}\mathrm{only}\mathrm{possible}$$ $$1+acosx-b=0\Rightarrow \lim \frac{x(1+acosx)-bsinx}{x^5}=1$$ $$\mathrm{and}\mathrm{it}\mathrm{is}\mathrm{impossible}$$ $$\Rightarrow \lim \frac{x(1+acosx)-bsinx}{x^5}=1\Rightarrow 1+acosx-b=0$$ $$\mathrm{it}\mathrm{is}\mathrm{also}\mathrm{like}\mathrm{as}1+\mathrm{acosx}-\mathrm{kb}=0$$ $$\Rightarrow \lim \frac{x(1+acosx)-bsinx}{x^5}=1$$ $$\mathrm{If}\left(1\right)\mathrm{is}\mathrm{true}\mathrm{then}1+\mathrm{acosx}-\mathrm{kb}=0$$ $$\mathrm{is}\mathrm{also}\mathrm{true}.\mathrm{Sir}\mathrm{should}\mathrm{see}\mathrm{once}\mathrm{again}$$ $${\mathrm{sir}}^{\prime }\mathrm{argument}$$

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