A) \[e\]
B) \[{{e}^{-1}}\]
C) \[{{e}^{-5}}\]
D) \[{{e}^{5}}\]
Correct Answer: C
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{x-3}{x+2} \right)}^{x}}=\underset{x\to \infty }{\mathop{\lim }}\,{{\left[ \frac{1-\frac{3}{x}}{1+\frac{2}{x}} \right]}^{x}}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{\left( 1-\frac{3}{x} \right)}^{x}}}{{{\left( 1+\frac{2}{x} \right)}^{x}}}\] \[=\frac{{{e}^{-3}}}{{{e}^{2}}}\] \[={{e}^{-5}}\]
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