A) \[\underset{x\to 1+}{\mathop{\lim }}\,f(x)=2\]
B) \[\underset{x\to 1-}{\mathop{\lim }}\,f(x)=3\]
C) \[f(x)\]is discontinuous at \[x=1\]
D) None of these
Correct Answer: C
Solution :
\[f(x)=\left\{ \frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1} \right\}\], for \[x\ne 1\] \[=2\,\], for \[x=1\] \[f(1)=2,\,\,f(1+)=\underset{x\to 1+}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1}=\underset{x\to 1+}{\mathop{\lim }}\,\,\frac{(x-3)}{(x+1)}=-1\] \[f(1-)=\underset{x\to 1-}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1}=-1\,\,\Rightarrow \,\,f(1)\ne f(1-)\] Hence the function is discontinuous at \[x=1.\]
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