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

$$\underset{x\to 0}{\lim }\frac{x(1+a\cos x)-b\sin x}{x^5}=1$$$$\mathrm{find}\mathrm{the}\mathrm{value}a\mathrm{and}b$$

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

$$\underset{x\to 0}{\lim }\frac{x+ax(1-\frac{x^2}{2!}+\frac{x^4}{4\cdot }-..)-b(x-\frac{x^3}{3!}+\frac{x^5}{5!}+..)}{x^5}$$$$\underset{x\to 0}{\lim }\frac{x(1+a-b)+a(\frac{-x^3}{2!}+\frac{x^5}{4!}+..)-b(\frac{-x^3}{3!}+\frac{x^6}{5!}+..)}{x^5}$$$$\underset{x\to 0}{\lim }\frac{1+a-b+a(-\frac{x^2}{2!}+\frac{x^4}{4!}+..)-b(\frac{-x^2}{3!}+\frac{x^5}{5!}..)}{x^4}$$$$1+a-b=0$$$$\underset{x\to 0}{\lim }\frac{a(\frac{-1}{2!}+\frac{x^2}{4!}+..)-b(\frac{-1}{3!}+\frac{x^3}{5!}+..)}{x^2}$$$$\frac{-a}{2}+\frac{b}{6}=0b=3a$$$$1+a-3a=0a=\frac{1}{2}b=\frac{3}{2}$$

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