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