A) 0
B) 1
C) \[-1\]
D) 2
E) does not exist
Correct Answer: A
Solution :
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-2x+1}{|{{x}^{2}}-1|}\] \[|{{x}^{2}}-1|=\left\{ \begin{matrix} -({{x}^{2}}-1), & if-1<x<1 \\ {{x}^{2}}-1, & if\,x>1 \\ \end{matrix} \right.\] \[\therefore \] \[\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\frac{{{x}^{2}}-2x+1}{{{x}^{2}}-1}=\underset{x\to 1}{\mathop{\lim }}\,\frac{2x-2}{2x}=0\] \[\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\frac{{{x}^{2}}-2x+1}{1-{{x}^{2}}}=\underset{x\to 1}{\mathop{\lim }}\,\frac{2x-2}{-2x}=0\] \[\therefore \] \[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-2x+1}{|{{x}^{2}}-1|}=0\]
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