A) \[{{e}^{4}}\]
B) \[{{e}^{2}}\]
C) \[{{e}^{3}}\]
D) \[e\]
Correct Answer: A
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,\left[ 1+\frac{4x+1}{{{x}^{2}}+x+2} \right]=\underset{x\to \infty }{\mathop{\lim }}\,{{[{{(1+\alpha )}^{1/\alpha }}]}^{\alpha x}}\] when \[\alpha =\frac{4x+1}{{{x}^{2}}+x+2}\] \[=\frac{4+\frac{1}{x}}{x\left( x+\frac{1}{x}+\frac{2}{{{x}^{2}}} \right)}\to 0\] as\[x\to \infty \] and \[\alpha x=\frac{4+\frac{1}{x}}{1+\frac{1}{x}+\frac{2}{{{x}^{2}}}}\to 4as\,x\to \infty \] Given limit \[={{e}^{4}}\]
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