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Let $\mathrm{f}$ be the function defined by $f(x)=\left\{\begin{array}{ll}\frac{x^{2}-1}{x^{2}-2|x-1|-1}, & x \neq 1 \\ 1 / 2, & x=1\end{array}\right.$
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The correct answer is:
The function is not continuous at $x=1$
For $x<1, f(x)=\frac{x^{2}-1}{x^{2}+2 x-3}=\frac{x+1}{x+3}$
$\therefore \lim _{x \rightarrow 1^{-}} \mathrm{f}(\mathrm{x})=\frac{1}{2}$
For $x>1, f(x)=\frac{x^{2}-1}{x^{2}-2 x+1}=\frac{x+1}{x-1}$
$\therefore \lim _{x \rightarrow 1^{+}} f(x)=\infty$
$\therefore$ The function is not continuous at $\mathrm{x}=1$.
$\therefore \lim _{x \rightarrow 1^{-}} \mathrm{f}(\mathrm{x})=\frac{1}{2}$
For $x>1, f(x)=\frac{x^{2}-1}{x^{2}-2 x+1}=\frac{x+1}{x-1}$
$\therefore \lim _{x \rightarrow 1^{+}} f(x)=\infty$
$\therefore$ The function is not continuous at $\mathrm{x}=1$.
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