If \(\begin{array}{*{20}{c}} {\lim }\\ {x \to 0} \end{array}\frac{{\left( {x\left( {1 + a\cos \left( x \right)} \right) - b\sin \left( x \right)} \right)}}{{{x^3}}} = 1\)

then the values of a and b are 

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = \frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\)

Calculation:

Given:

\(\begin{array}{*{20}{c}} {\lim }\\ {x \to 0} \end{array}\frac{{\left( {x\left( {1 + a\cos \left( x \right)} \right) - b\sin \left( x \right)} \right)}}{{{x^3}}} = 1\)

\(\begin{array}{*{20}{c}} {\lim }\\ {x \to 0} \end{array}\frac{{\left( {x\left( {1 + a\cos \left( x \right)} \right) - b\sin \left( x \right)} \right)}}{{{x^3}}} \) is 0/0 form

Applying L-Hospital rule:

\(\displaystyle \lim_{x \to 0} \frac{\frac{d}{dx}[x(1~+~acosx~-~bsinx)]}{\frac{d}{dx}(x^3)}\)

⇒ \(\displaystyle \lim_{x \to 0} \frac{[(1~+~acosx~-ax(sinx)-~bcosx)]}{(3x^2)}\)

⇒ \(\displaystyle \lim_{x \to 0} \frac{[(1~+~(a~-~b)cosx~-x(asinx)]}{(3x^2)}~=~1\)

To satisfy limit, 1 + a - b = 0

⇒ a + 1 = b ................................................................ (1)

Applying L-Hospital rule again:

\(\displaystyle \lim_{x \to 0} \frac{[(-(a~-~b)sinx~-~asinx~-~x(acosx)]}{(6x)}~=~1\)

⇒ \(\displaystyle \lim_{x \to 0} \frac{(b~-~2a)}{6}\frac{sinx}{x}~-\displaystyle \lim_{x \to 0}~\frac{a}{6}cosx~=~1\)

To satisfy limit, \(\frac{(b~-~2a)}{6}~-~\frac{a}{6}~=~1\) ...............................(2)

Solving (1) and (2)

a = \(\frac{-5}{2}\) , b = \(\frac{-3}{2}\)