If $\lim_{x\to 0}\frac1{x^3}\left(\frac1{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)=l$, then what is the value of $\frac{1}{a}-\frac{2}{l}+\frac{3}{b}$? If the function 
$$\lim_{x\to 0}\frac1{x^3}\left(\frac1{\sqrt{1+x}}-\frac{1+ax}{1+bx}\right)$$ exists and has a value equal to $l$ then what will be the value of $\frac{1}{a}-\frac{2}{l}+\frac{3}{b}$

You can write the limit as
$$
\lim_{x\to0}\frac{1+bx-(1+ax)\sqrt{1+x}}{x^3}\frac{1}{\sqrt{1+x}(1+bx)}
$$
Since the second factor has limit $1$, we can disregard it. Use Taylor expansion up to order $3$:
$$
\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+o(x^3)
$$
Then the numerator is
$$
1+bx-1-\frac{1}{2}x+\frac{1}{8}x^2-\frac{1}{16}x^3
-ax-\frac{a}{2}x^2+\frac{a}{8}x^3+o(x^3)\\=
\left(b-\frac{1}{2}-a\right)x+
\left(\frac{1}{8}-\frac{a}{2}\right)x^2+
\left(-\frac{1}{16}+\frac{a}{8}\right)x^3+o(x^3)
$$
In order the limit is finite, we need
\begin{cases}
b-\dfrac{1}{2}-a=0
\\[4px]
\dfrac{1}{8}-\dfrac{a}{2}=0
\end{cases}
and the limit is
$$
l=-\frac{1}{16}+\frac{a}{8}
$$
