Another limit

$\begin{aligned} l & = \lim_{x \to 0} \frac 1{x^3} \left(\frac 1{\sqrt{1+x}} - \frac {1+ax}{1+bx}\right) & \small \color{#3D99F6} \text{A 0/0 case, L'Hôpital's rule applies} \\ & = \lim_{x \to 0} \frac 1{3x^2} \left(- \frac 12 (1+x)^{-\frac 32} - \frac {a(1+bx)-b(1+ax)}{(1+bx)^2}\right) & \small \color{#3D99F6} \text{Differentiate up and down w,r,t, }x \\ & = \lim_{x \to 0} \frac 1{3x^2} \left(- \frac 12 (1+x)^{-\frac 32} - \frac {a-b}{(1+bx)^2}\right) & \small \color{#3D99F6} \text{For }l \text{ to exist }- \frac 12 - a + b = 0 \\ & = \lim_{x \to 0} - \frac 1{6x^2} \left((1+x)^{-\frac 32} - \frac 1{(1+bx)^2}\right) & \small \color{#3D99F6} \implies a-b = - \frac 12 \\ & = \lim_{x \to 0} - \frac 1{12x} \left(-\frac 32(1+x)^{-\frac 52} + \frac {2b}{(1+bx)^3}\right) & \small \color{#3D99F6} \text{Apply L'Hôpital's rule again} \\ & = \lim_{x \to 0} \frac 1{8x} \left((1+x)^{-\frac 52} - \frac 1{\left(1+\frac 34 x\right)^3}\right) & \small \color{#3D99F6} \text{For }l \text{ to exist } \implies b = \frac 34 \\ & = \lim_{x \to 0} \frac 1{8} \left(-\frac 52(1+x)^{-\frac 72} + \frac {3\times \frac 34}{\left(1+\frac 34 x\right)^4}\right) & \small \color{#3D99F6} \text{Apply L'Hôpital's rule again} \\ & = \frac 1{8} \left(-\frac 52 + \frac 94 \right) = - \frac 1{32} \end{aligned}$

Therefore, $\dfrac 1a - \dfrac 2l + \dfrac 3b = 4+64+4 = \boxed{72}$ .

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Alternative solution as suggested by @Aaghaz Mahajan

$\begin{aligned} l & = \lim_{x \to 0} \frac 1{x^3} \left({\color{#3D99F6} \frac 1{\sqrt{1+x}}} - \frac {1+ax}{\color{#3D99F6} 1+bx}\right) & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0} \frac 1{x^3} \left({\color{#3D99F6}\left(1-\frac 12 x + \frac 38 x^2 - \frac 5{16}x^3 + \cdots \right)} - (1+ax)\color{#3D99F6}\left(1 - bx + b^2x^2 - b^3x^3 + \cdots \right) \right) \\ & = \lim_{x \to 0} \frac 1{x^3} \left(\left(1-\frac 12 x + \frac 38 x^2 - \frac 5{16}x^3 + \cdots \right) - \left(1 + (a- b)x + (b^2-ab)x^2 + (ab^2- b^3) x^3 + \cdots \right) \right) \\ & = \lim_{x \to 0} \left(\left(b-a-\frac 12\right) \frac 1{x^2} + \left(\frac 38 +ab - b^2 \right) \frac 1x - \left(\frac 5{16} + ab^2 - b^3 \right) + O(x) \right) \end{aligned}$

For $l$ to exist,

$\implies \begin{cases} b-a-\dfrac 12 = 0 \\ \dfrac 38 +ab - b^2 = \dfrac 38 +b(a-b) = \dfrac 38 - \dfrac b2 = 0 \end{cases} \implies b = \dfrac 34 \implies a = \dfrac 14$

Then $l = - \left(\dfrac 5{16} + ab^2 - b^3 \right) = - \dfrac 5{16} - \dfrac 14 \times \dfrac 9{16} + \dfrac {27}{64} = - \dfrac 1{32}$ and $\dfrac 1a - \dfrac 2l + \dfrac 34 = 4 + 64 + 4 = \boxed{72}$ .