Answer
0
Work Step by Step
1. f is a closed function (we know by Th.10.1 that it is continuous), that is, L= $\displaystyle \lim_{x\rightarrow a}f(x)$ = $f(a)$,
for all a from the domain of f.
2. evaluating: $f(1)$, (plugging $x=1$) results in an indeterminate form 0/0.
So, we see that $x=1$ is NOT in the domain of f.
We apply case 2 (of step 2). Simplifying $f(x)$,
$\displaystyle \frac{x^{2}-2x+1}{x^{2}-x}=$... recognize a square of a difference
$=\displaystyle \frac{(x-1)^{2}}{x(x-1)}=$...reduce...$=\displaystyle \frac{x-1}{x}=g(x)$
$g(1)=0$
($1$ is in the domain of g, g is closed-form), so
$\displaystyle \lim_{x\rightarrow 1}f(x)=\lim_{x\rightarrow 1}g(x)=g(1)=0$
