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Q. $ \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{2}}-2x+1}{|{{x}^{2}}-1|} $ is equal to:

KEAMKEAM 2000

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-11 \\ \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 $