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