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

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

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

find the value a and b

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

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

\lim_{x \to 0} \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}}

\lim_{x \to 0} \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

\lim_{x \to 0} \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} = 0 b = 3 a

1 + a - 3 a = 0 a = \frac{1}{2} b = \frac{3}{2}