A) \[\underset{x\to 1}{\mathop{\lim }}\,f(x)=f(1)\]
B) \[\underset{x\to 1}{\mathop{\lim }}\,f(x)=3\]
C) \[\underset{x\to 1}{\mathop{\lim }}\,f(x)=2\]
D) \[\underset{x\to 1}{\mathop{\lim }}\,f(x)\]does not exist
Correct Answer: D
Solution :
L.H.L.\[=\underset{x\to 1-0}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,\,\,(1-h)=\underset{h\to 0}{\mathop{\lim }}\,\,\,3(1-h)\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,(3-3h)=3-3\,.\,0=3\] R.H.L.\[=\underset{x\to 1+0}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim \,\,}}\,\,f\,(1+h)=\underset{h\to 0}{\mathop{\lim }}\,\,\,[5-3(1+h)]\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\,(2-3h)=2-3\,.\,0=2\] Hence \[\underset{x\to 1}{\mathop{\lim }}\,\,\,f(x)\] does not exist.
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