<s>[INST] 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 $\frac{1}{a}-\frac{2}{l}+\frac{3}{b}$
 [/INST] Let $$f(x)=\frac{1}{\sqrt{1+x}}-\frac{1+ax}{1+bx}$$ and let $$g(x)=x^3.$$ Apply L'Hospital's rule.
First derivatives
We have $$f'(x)=-\frac{1}{2(1+x)^{3/2}}-\frac{a}{1+bx}-\frac{b(1+ax)}{(1+bx)^2}$$ so that $f'(0)=b-a-1/2$. Also $g'(x)=3x^2$ so that $g'(0)=0$. Thus since there is a finite limit $l$ it must be that $f'(0)=0$ i.e. $b=a+1/2$. Substituting this into $f'(x)$ gives $$f'(x)=\frac{1}{2(1+x)^{3/2}}-\frac{2}{[2+(1+2a)x]^2}.$$ 
Second derivatives
Differentiating again, we have $$f''(x)=\frac{3}{4(x+1)^{5/2}}-\frac{4(2a+1)}{[2+(1+2a)x]^3},$$ so that $f''(0)=1/4-a$. Since $g''(x)=6x$ so that $g''(0)=0$, we need $f''(0)=0$ which gives $a=1/4$. Substituting this value gives $$f''(x)=\frac{3}{4(x+1)^{5/2}}-\frac{48}{(3x+4)^3}.$$
Third derivatives
Differentiating again, we have $$f'''(x)=-\frac{15}{8(x+1)^{7/2}}+\frac{432}{(3x+4)^4}.$$ Thus $f'''(0)=-3/16$. Finally, since $g'''(x)=6$, the limit is
$$l=\frac{f'''(0)}{g'''(0)}=\frac{(-3/16)}{6}=-\frac{1}{32}.$$
Conclusion
Thus $$\frac{1}{a}-\frac{2}{l}+\frac{3}{b}=4+64+4=72.$$
(Probably there is a faster way...)
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