A) \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=1\]
B) \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=0\]
C) at\[x=0,f(x)\]is continuous
D) \[\underset{x\to 0}{\mathop{\lim }}\,f(x)\]does not exist
Correct Answer: B
Solution :
\[f(x)=\left\{ \begin{matrix} x, & x<0 \\ 1, & x=0 \\ {{x}^{2}}, & x>0 \\ \end{matrix} \right.\] At\[x<0,\] \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=0\] At\[x>0,\] \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=0\] \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=0\]
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