A) \[{{e}^{2}}\]
B) \[{{e}^{3}}\]
C) e
D) \[{{e}^{-1}}\]
Correct Answer: A
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{x+3}{x+1} \right)}^{x+1}}\]\[=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{2}{x+1} \right)}^{\frac{x+1}{2}.2}}\] \[={{\left\{ \underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{2}{x+1} \right)}^{\frac{x+1}{2}}} \right\}}^{2}}\]\[={{e}^{2}}\].
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