Define a function f by f(x)=x+2ln(x) and let g(x)=f(x)-A-B(x-1)-C(x-1)^2. Find the values of A, B, and C so that L=lim x→1 g(x)/(x-1)^3 exists and is finite.

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- Nov 3rd 2009, 03:10 PMVelvet LoveSolving
Define a function f by f(x)=x+2ln(x) and let g(x)=f(x)-A-B(x-1)-C(x-1)^2. Find the values of A, B, and C so that L=lim x→1 g(x)/(x-1)^3 exists and is finite.

- Nov 4th 2009, 06:47 PMMedia_ManL'Hospital
L'Hospital's Rule states that a limit $\displaystyle L=\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$ if the limit is indeterminate, that is, $\displaystyle \frac{f(a)}{g(a)}=\frac00$

Here, if we substitute x=1 into the function $\displaystyle \frac{x-\ln x-A-B(x-1)-C(x-1)^2}{(x-1)^3}$, we get $\displaystyle \frac{1-A}{0}$. Notice that if the numerator is NOT zero, then the limit will run off to $\displaystyle \pm\infty$. Therefore, it is necessary that A=1 so the limit is indeterminate, otherwise the limit will not converge at all. Apply L'Hospital's Rule...

Now $\displaystyle L=\lim_{x\to1}\frac{1+2/x-B-2C(x-1)}{3(x-1)^2}$, so repeat the process to find B. Keep going and you can get A,B,C, and also the final value of the limit L.