A) \[e\]
B) \[{{e}^{2}}\]
C) \[{{e}^{3}}\]
D) \[{{e}^{4}}\]
Correct Answer: D
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( \frac{{{x}^{2}}+5x+3}{{{x}^{2}}+x+2} \right)}^{x}}\] \[=\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{4x+1}{{{x}^{2}}+x+2} \right)}^{x\times \frac{{{x}^{2}}+x+2}{4x+1}\times \frac{4x+1}{{{x}^{2}}+x+2}}}\] \[={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\frac{4{{x}^{2}}+x}{{{x}^{2}}+x+2}}}\] \[={{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\frac{4+\frac{1}{x}}{1+\frac{1}{x}+\frac{2}{{{x}^{2}}}}}}={{e}^{4}}\]
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