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 July 8th, 2011, 03:54 AM #1 Senior Member   Joined: Jul 2011 Posts: 395 Thanks: 15 limit question $\displaystyle \lim_{x\rightarrow 0}\frac{\left\{\frac{1}{\sqrt{1+x^2}}-\frac{(1+ax)}{(1+bx)}\right\}}{x^3}= L$ find value of $a,b,L$
 July 8th, 2011, 06:49 AM #2 Senior Member   Joined: Jul 2011 Posts: 395 Thanks: 15 Re: limit question $\text\lim_{x\rightarrow 0}\; \; \frac{1}{x^3}\left((1+x^2)^{-\frac{1}{2}}-(1+ax)(1+bx)^{-1}\right)$ using $\text (1+x)^n=1+nx+\frac{n(n-1).x^2}{2!}+\frac{n(n-1)(n-2).x^3}{3!}+......................$ $\text\lim_{x\rightarrow 0}\frac{1}{x^3}.\left[(1-\frac{1}{2}.x^2+\frac{1}{2}.\frac{3}{2}.\frac{x^4} {2!}+....) \;-\;\left((1+ax).(1-bx+b^2x^2-b^3x^3+.....+)\right)\right]$ $\text\lim_{x\rightarrow 0}\frac{1}{x^3}.\left[1-\frac{x^2}{2}+\frac{3}{8}x^4 -\; 1-ax+bx+abx^2-b^2x^2-ab^2x^3+b^3x^3+ab^4x^4\right]$ $\text\lim_{x\rightarrow 0}\frac{1}{x^3}.\left[(a-b)x+(ab-b^2-\frac{1}{2})x^2+(b^3-ab^2)x^3+......\right]$ now for a limit to have finite, we have $\text (a-b)=0$ and $\text (ab-b^2-\frac{1}{2})=0$ but I am getting $\text a=b$ and $\text -\frac{1}{2}=0$ so where i am doing wrong plz explain me thanking you
 July 9th, 2011, 01:30 AM #3 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Re: limit question Well using l'Hôpital's rule isn't that easy...

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