# Solving

L'Hospital's Rule states that a limit $L=\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$ if the limit is indeterminate, that is, $\frac{f(a)}{g(a)}=\frac00$
Here, if we substitute x=1 into the function $\frac{x-\ln x-A-B(x-1)-C(x-1)^2}{(x-1)^3}$, we get $\frac{1-A}{0}$. Notice that if the numerator is NOT zero, then the limit will run off to $\pm\infty$. 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 $L=\lim_{x\to1}\frac{1+2/x-B-2C(x-1)}{3(x-1)^2}$, 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.