If \[f(x)=\left\{ \begin{align} & {{(1+2x)}^{1/x}},\,\text{for }x\ne 0 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{e}^{2}},\,\text{for }x=0\,\,\, \\ \end{align} \right.\], then

\[f(x)\]is discontinuous at \[x=0\]

\[\underset{x\to 0+}{\mathop{\lim }}\,f(x)=e\]

\[\underset{x\to 0-}{\mathop{\lim }}\,f(x)={{e}^{2}}\]

If \[f(x)=\left\{ \begin{align} & {{(1+2x)}^{1/x}},\,\text{for }x\n

1.	If \[f(x)=\left\{ \begin{align} & {{(1+2x)}^{1/x}},\,\text{for }x\ne 0 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{e}^{2}},\,\text{for }x=0\,\,\, \\ \end{align} \right.\], then
A.	\[\underset{x\to 0+}{\mathop{\lim }}\,f(x)=e\]
B.	\[\underset{x\to 0-}{\mathop{\lim }}\,f(x)={{e}^{2}}\]
C.	\[f(x)\]is discontinuous at \[x=0\]
D.	None of these
Answer» C. \[f(x)\]is discontinuous at \[x=0\]

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If \[f(x)=\left\{ \begin{align} & {{(1+2x)}^{1/x}},\,\text{for }x\ne 0 \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{e}^{2}},\,\text{for }x=0\,\,\, \\ \end{align} \right.\], then

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