Answer
L'Hospital's rule applies,
limit = $0$
Work Step by Step
Theorem 11.3 L'Hospital's Rule:
If $f$ and $g$ are two differentiable functions such that substituting $x=a$ in the expression $\displaystyle \frac{f(x)}{g(x)}$ gives the indeterminate form $\displaystyle \frac{0}{0}$ or $\displaystyle \frac{\infty}{\infty}$, then
$\displaystyle \lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a} \displaystyle \frac{f^{\prime}(x)}{\mathrm{g}^{\prime}(x)}$.
That is, we can replace $f(x)$ and $g(x)$ with their derivatives and try again to take the limit.
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When x approaches 1,
the numerator approaches 0,
the denominator approaches 0,
$\displaystyle \frac{f(x)}{g(x)}$ gives the indeterminate form $\displaystyle \frac{0}{0}$,
L'Hospital's rule applies.
$\displaystyle \lim_{x\rightarrow 1}\frac{x^{2}-2x+1}{x^{2}-x}=\lim_{x\rightarrow 1} \displaystyle \frac{[x^{2}-2x+1]^{\prime}}{[x^{2}-x]^{\prime}}$
$=\displaystyle \lim_{x\rightarrow 1} \displaystyle \frac{2x-2}{2x-1}=\frac{2-2}{2-1}=0$
