A) \[\underset{x\to 2+}{\mathop{\lim }}\,f(x)\ne 0\]
B) \[\underset{x\to 2-}{\mathop{\lim }}\,f(x)\ne 0\]
C) \[\underset{x\to 2+}{\mathop{\lim }}\,f(x)\ne \underset{x\to 2-}{\mathop{\lim }}\,f(x)\]
D) \[f(x)\]is continuous at \[x=2\]
Correct Answer: D
Solution :
Here \[f(2)=0\] \[\underset{x\to 2-}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(2-h)=\underset{h\to 0}{\mathop{\lim }}\,\,\,|\,\,2-h-2\,\,|=0\] \[\underset{x\to 2-}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,f(2-h)=\underset{h\to 0}{\mathop{\lim }}\,\,\,|\,\,2-h-2\,\,|=0\] Hence it is continuous at \[x=2\].
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