Answer: Start with, $lim_{x \to 0 } \dfrac {x(1+acosx)-bsinx}{x^3}=1 \Rightarrow lim_{x \to 0 } \dfrac {x[1+a(1-x^2/2!+x^4/4!-...)]-b[x-x^3/3!+x^5/5!-...]}{x^3}=1$

$=lim_{x \to 0 } \dfrac {[x+ax-ax^3/2!+ax^5/4!-...)]-[bx-bx^3/3!+bx^5/5!-...]}{x^3}=1$

Or,

$=lim_{x \to 0 } [(1+a-b)x/x^3+(b/3!-a/2!)x^3/x^3+(a/4!-b/5!)x^5/x^3...)]=1$

From the second term,

$b/6-a/2=1 \Rightarrow b-3a=6.$..................(1)

And from the first term,

$1+a-b=0$................................(2).

Solving equations (1) and (2), we get

$1+a-3a-6=0 \Rightarrow -2a-5=0 \Rightarrow a=-5/2.$     

Now putting the value of $a$ in (1), we get $b=-15/2+6=-3/2.$ Thus $a=-5/2,b=-3/2.$            Answer.