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