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by **stuart clark** on Fri Jul 08, 2011 6:17 pm

$\displaystyle \lim_{x\rightarrow 0}\,\frac{1}{x^3}.\left(\frac{1}{\sqrt{x^2+1}}\;-\;\frac{1+ax}{1+bx}\right) = L$ (finite quantity)

then $a,b$ and $L$ is

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by **nona.m.nona** on Fri Jul 08, 2011 9:08 pm

Have you combined the expression into one term? If so, where did this lead? Where are you bogging down?

Please be complete. Thank you.

by **gliexon** on Mon Jul 18, 2011 9:30 pm

I had 0/0 so I derived 1/x^3 and (1/sqrt[x^2+1] - (1+ax)/(1+bx) separately, simplified the equation and reached 1/3 lim x=>0 (b-a)/x^2, but if L is finite I don't really know where to go from there. Did I do something wrong?

by **Martingale** on Tue Jul 19, 2011 3:10 am

it doesn't look possible. I looked at the Taylor expansion of $\frac{1}{\sqrt{x^2+1}}\;-\;\frac{1+ax}{1+bx}$ . Even with the best choice of a,b the expansion includes an $x^2$ term and therefore the limit would not exist

The Purplemath Forums

limit, x=>0, (1/x^3)(1/sqrt[x^2+1] - (1+ax)/(1+bx)) = l

limit, x=>0, (1/x^3)(1/sqrt[x^2+1] - (1+ax)/(1+bx)) = l

Re: limit, x=>0, (1/x^3)(1/sqrt[x^2+1] - (1+ax)/(1+bx)) = l

Re: limit, x=>0, (1/x^3)(1/sqrt[x^2+1] - (1+ax)/(1+bx)) = l

Re: limit, x=>0, (1/x^3)(1/sqrt[x^2+1] - (1+ax)/(1+bx)) = l