If \[f(x)=\left\{ \begin{align} & \frac{{{x}^{2}}-1}{x+1},\,\text{when }x\ne -1 \\ & \,\,\,\,\,\,\,\,-2,\,\text{when }x=-1 \\ \end{align} \right.\],then

\[\underset{x\to {{(-1)}^{-}}}{\mathop{\lim }}\,f(x)=-2\]

\[\underset{x\to {{(-1)}^{+}}}{\mathop{\lim }}\,f(x)=-2\]

\[f(x)\]is continuous at \[x=-1\]

If \[f(x)=\left\{ \begin{align} & \frac{{{x}^{2}}-1}{x+1},\,\text{w

1.	If \[f(x)=\left\{ \begin{align} & \frac{{{x}^{2}}-1}{x+1},\,\text{when }x\ne -1 \\ & \,\,\,\,\,\,\,\,-2,\,\text{when }x=-1 \\ \end{align} \right.\],then
A.	\[\underset{x\to {{(-1)}^{-}}}{\mathop{\lim }}\,f(x)=-2\]
B.	\[\underset{x\to {{(-1)}^{+}}}{\mathop{\lim }}\,f(x)=-2\]
C.	\[f(x)\]is continuous at \[x=-1\]
D.	All the above are correct
Answer» E.

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If \[f(x)=\left\{ \begin{align} & \frac{{{x}^{2}}-1}{x+1},\,\text{when }x\ne -1 \\ & \,\,\,\,\,\,\,\,-2,\,\text{when }x=-1 \\ \end{align} \right.\],then

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