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Question Number 38893 by gunawan last updated on 01/Jul/18

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

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

$$\underset{x\to 0}{\lim }\frac{ax+bx(1-\frac{x^2}{2!}+\frac{x^4}{4!}+...)-c(\frac{x}{1}-\frac{x^3}{3!}+\frac{x^5}{5!}+..)}{x^5}$$$$=\underset{x\to 0}{\lim }\frac{x(a+b-c)+b(\frac{-x^3}{2!}+\frac{x^5}{4!}+..)-c(-\frac{x^3}{3!}+\frac{x^5}{5!}..)}{x^5}$$$$\frac{lim}{x\to 0}\frac{(a+b-c)+b(-\frac{3x^2}{2!}+\frac{5x^4}{4!}+..)-c(\frac{-3x^2}{3!}+\frac{5x^4}{5!}+..}{5x^4}$$$$sofor\left(\frac{0}{0}\right)forma+b-c=0.....eqn1$$$$=\underset{x\to 0}{\lim }\frac{b(-\frac{3}{2!}+\frac{5x^2}{4!}+..)-c(\frac{-3}{3!}+\frac{5x^2}{5!}+..)}{5x^2}$$$$b(-\frac{3}{2!})+c\left(\frac{3}{3!}\right)=0for\left(\frac{0}{0}\right)form$$$$\frac{c}{2}-\frac{3b}{2}=0c=3b$$$$\underset{x\to 0}{\lim }\frac{b(\frac{5x^2}{4!}+..)-c(\frac{5x^2}{5!}+..)}{5x^2}=1$$$$\frac{b}{4!}-\frac{c}{5!}=1\frac{b}{24}-\frac{c}{120}=1$$$$\frac{b}{24}-\frac{3b}{120}=1$$$$\frac{b}{24}-\frac{b}{40}=1so\frac{5b-3b}{3\times 8\times 5}=1$$$$2b=3\times 8\times 5b=60$$$$c=3\times 60=180$$$$a+b-c=0a=c-b$$$$a=180-60=120$$$$$$

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