Determine a and b in order that _(x→0) ^(lim) ((x(1+a cos x)−b sin x )/x^3 )=1.


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Question Number 84065 by niroj last updated on 09/Mar/20

$Determine a and b in order that$ $\underset{x \to 0}{\overset{\lim}{}} \frac{x (1 + a \cos x) - b \sin x}{x^{3}} = 1 .$
Commented by mathmax by abdo last updated on 09/Mar/20

$\Rightarrow {l i m}_{x \to 0} \frac{x (1 + a c o s x) - b s i n x}{x^{3}} - 1 = 0 \Rightarrow$ ${l i m}_{x \to 0} \frac{x (1 + a c o s x) - b s i n x - x^{3}}{x^{3}} = 0 \Rightarrow$ $u (o) = u^{^{'}} (0) = u^{(2)} (0) = u^{(3)} (0) = 0 a n d v (0) = v^{^{'}} (0) = v^{(2)} (0) = v^{(3)} (0) = 0 w i t h$ $u (x) = x (1 + a c o s x) - b s i n x - x^{3}$ $u (0) = 0 \Rightarrow 0 = 0 (v)$ $u^{^{'}} (x) = 1 + a c o s x + x (- a s i n x) - b c o s x - 3 x^{2}$ $u^{^{'}} (0) = 0 \Rightarrow 1 + a - b = 0 \Rightarrow b = 1 + a$ $u^{(2)} (x) = - a s i n x - a s i n x + x (- a c o s x) + b s i n x - 6 x$ $u^{(2)} (0) = 0 \Rightarrow 0 = 0 (v)$ $u^{(3)} (x) = - 2 a c o s x - a c o s x + a x s i n x + b c o s x - 6$ $u^{(3)} (0) = 0 \Rightarrow - 3 a + b - 6 = 0 \Rightarrow b = 3 a + 6 \Rightarrow 1 + a = 3 a + 6 \Rightarrow 2 a + 6 = 1 \Rightarrow$ $a = - \frac{5}{2} a n d b = - \frac{3}{2} (f o r v t h e c o n d i t i o n s i s v e r i f i e d . .)$
Commented by mathmax by abdo last updated on 09/Mar/20

$v (x) = x^{3}$
	Answered by TANMAY PANACEA last updated on 09/Mar/20
	$lim_(x→0) ((x{1+a(1−(x^2 /(2!))+(x^4 /(4!))+..)}−b(x−(x^3 /(3!))+(x^5 /(5!))+..))/x^3 ) lim_(x→0) ((x+ax−((ax^3 )/(2!))+((ax^5 )/(4!))−bx+((bx^3 )/(3!))−((bx^5 )/(5!)))/x^3 ) lim_(x→0) ((x(1+a−b)+x^3 ((b/6)−(a/2))+others terms)/x^3 ) 1+a−b=0 (b/6)−(a/2)=1 b−3a=6 1+a−3a=6 −2a=5→a=((−5)/2) b=1+a b=1−(5/2)=((−3)/2)$
	$\lim_{x \to 0} \frac{x {1 + a (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + . .)} - b (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + . .)}{x^{3}}$ $\lim_{x \to 0} \frac{x + a x - \frac{{a x}^{3}}{2!} + \frac{{a x}^{5}}{4!} - b x + \frac{{b x}^{3}}{3!} - \frac{{b x}^{5}}{5!}}{x^{3}}$ $\lim_{x \to 0} \frac{x (1 + a - b) + x^{3} (\frac{b}{6} - \frac{a}{2}) + o t h e r s t e r m s}{x^{3}}$ $1 + a - b = 0$ $\frac{b}{6} - \frac{a}{2} = 1$ $b - 3 a = 6$ $1 + a - 3 a = 6$ $- 2 a = 5 \to a = \frac{- 5}{2}$ $b = 1 + a$ $b = 1 - \frac{5}{2} = \frac{- 3}{2}$
	Commented by niroj last updated on 09/Mar/20

	$e x e l l e n t d e a r . t h a n k y o u s o m u c h .$
	Commented by TANMAY PANACEA last updated on 09/Mar/20

	$m o s t w e l c o m e s i r$
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