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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 $\frac{1}{a}-\frac{2}{l}+\frac{3}{b}$

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  • $\begingroup$ what your thinking? $\endgroup$
    – Empty
    Jul 15, 2016 at 15:44
  • $\begingroup$ Apply L'Hospital $\endgroup$
    – Empty
    Jul 15, 2016 at 15:44

2 Answers 2

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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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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 $$ l=-\frac{1}{16}+\frac{a}{8} $$

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