lim-x-0-x-a-b-cos-x-c-sin-x-x-5-1-find-the-value-a-b-and-c-

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

\lim_{x \to 0} \frac{x (a + b \cos x) - c \sin x}{x^{5}} = 1

find the value a, b, and c

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

\lim_{x \to 0} \frac{a x + b x (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \dots) - c (\frac{x}{1} - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + . .)}{x^{5}}

= \lim_{x \to 0} \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{l i m}{x \to 0} \frac{(a + b - c) + b (- \frac{3 x^{2}}{2!} + \frac{5 x^{4}}{4!} + . .) - c (\frac{- 3 x^{2}}{3!} + \frac{5 x^{4}}{5!} + . .}{5 x^{4}}

s o f o r (\frac{0}{0}) f o r m a + b - c = 0 \dots . . e q n 1

= \lim_{x \to 0} \frac{b (- \frac{3}{2!} + \frac{5 x^{2}}{4!} + . .) - c (\frac{- 3}{3!} + \frac{5 x^{2}}{5!} + . .)}{5 x^{2}}

b (- \frac{3}{2!}) + c (\frac{3}{3!}) = 0 f o r (\frac{0}{0}) f o r m

\frac{c}{2} - \frac{3 b}{2} = 0 c = 3 b

\lim_{x \to 0} \frac{b (\frac{5 x^{2}}{4!} + . .) - c (\frac{5 x^{2}}{5!} + . .)}{5 x^{2}} = 1

\frac{b}{4!} - \frac{c}{5!} = 1 \frac{b}{24} - \frac{c}{120} = 1

\frac{b}{24} - \frac{3 b}{120} = 1

\frac{b}{24} - \frac{b}{40} = 1 s o \frac{5 b - 3 b}{3 \times 8 \times 5} = 1

2 b = 3 \times 8 \times 5 b = 60

c = 3 \times 60 = 180

a + b - c = 0 a = c - b

a = 180 - 60 = 120