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