A) e
B) \[{{e}^{-1}}\]
C) \[{{e}^{-5}}\]
D) \[{{e}^{5}}\]
Correct Answer: C
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{x+2-5}{x+2} \right)}^{x}}=\underset{x\to \infty }{\mathop{\lim }}\,{{\left[ {{\left( 1-\frac{5}{x+2} \right)}^{\frac{x+2}{-5}}} \right]}^{-\,\frac{5x}{x+2}}}={{e}^{-5}}\] \[\left( \because \,\underset{x\to \infty }{\mathop{\lim }}\,\frac{-5x}{x+2}=\,\underset{x\to \infty }{\mathop{\lim }}\,\frac{-5}{1+\frac{2}{x}}=-5 \right)\].
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