limit with variables

$\begin{aligned} L & = \lim_{x \to 0} \frac {x^3}{(bx - {\color{#3D99F6}\sin x})\sqrt{a+x}} & \small \color{#3D99F6} \text{Using Maclaurin's series} \\ & = \lim_{x \to 0} \frac {x^3}{\left(bx - {\color{#3D99F6}\left(x - \frac {x^3}{3!}+\frac {x^5}{51}-\frac {x^7}{7!}+\cdots \right)}\right)\sqrt{a+x}} & \small \color{#3D99F6} \text{Setting } b = 1 \\ & = \lim_{x \to 0} \frac {x^3}{\left(\frac {x^3}{3!} - \frac {x^5}{51}+\frac {x^7}{7!}-\cdots \right)\sqrt{a+x}} & \small \color{#3D99F6} \text{Dividing up and down by }x^3 \\ & = \lim_{x \to 0} \frac 1{\left(\frac 1{3!} - \frac {x^2}{51}+\frac {x^4}{7!}-\cdots \right)\sqrt{a+x}} \\ & = \frac 1{\frac {\sqrt a}{3!}} = 1 \end{aligned}$

$\implies \dfrac {\sqrt a}{3!} = 1 \implies a = (3!)^2 = 36$

Therefore, $a+b = 36+1=\boxed{37}$ .