Limits

Since the given limit is of $\left[ \frac { 0 }{ 0 } \right]$ form;

By L Hopital's Rule,

$\underset { x\rightarrow 0 }{ lim } \frac { x(1+a\cos { x } )-b\sin { x } }{ { x }^{ 3 } }$ = $\underset { x\rightarrow 0 }{ lim } \frac { x(-a\sin { x } )+(1+a\cos { x } )-b\cos { x } }{ { 3x }^{ 2 } }$ ...........(1)

As $x\rightarrow 0$ , the denominator tends to 0 and the numerator becomes $1+a-b$ . Since the limit is finite and is equal to 1 as given in the question, the numerator must also tend to zero.

$\therefore \quad 1+a-b=0$

$\therefore \quad a-b=-1$ .........(2)

Applying L Hopital's Rule to equation (1) again;

$\underset { x\rightarrow 0 }{ lim } \frac { -ax\cos { x+(b-2a)\sin { x } \quad } }{ { 6x } }$

Again applying L.H. Rule and equating the limit with 1;

$\therefore \quad \underset { x\rightarrow 0 }{ lim } \frac { b-3a }{ 6 } =1$

$\therefore \quad b-3a=6$ ...........(3)

solving equations (2) and (3), we get:

$a=-\frac { 5 }{ 2 } \quad \& \quad b=-\frac { 3 }{ 2 }$

$\boxed{\therefore \quad a+b=-4}$