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

Limits, differentiation, related rates, integration, trig integrals, etc.

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

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

then $a,b$ and $L$ is
stuart clark

Posts: 2
Joined: Wed Jan 19, 2011 7:11 am

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

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

nona.m.nona

Posts: 249
Joined: Sun Dec 14, 2008 11:07 pm

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

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?
gliexon

Posts: 1
Joined: Mon Jul 18, 2011 8:59 pm

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

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

Martingale

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Location: USA