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

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

Postby stuart clark » Fri Jul 08, 2011 6:17 pm

(finite quantity)

then and is

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

Postby nona.m.nona » 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.

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

Postby gliexon » 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?

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

Postby Martingale » Tue Jul 19, 2011 3:10 am

it doesn't look possible. I looked at the Taylor expansion of . Even with the best choice of a,b the expansion includes an term and therefore the limit would not exist


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