A) \[\frac{2}{3}\]
B) \[\frac{3}{2}\]
C) 2
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
\[L=\underset{x\to \infty }{\mathop{Lim}}\,{{\left( 1+\frac{a}{x}-\frac{4}{{{x}^{2}}} \right)}^{2x}}\]must be of the from \[{{1}^{\infty }}\] \[L={{e}^{\underset{x\to \infty }{\mathop{lim}}\,\left( \frac{a}{x}-\frac{4}{{{x}^{2}}} \right)2x}}\]\[\Rightarrow \]\[L={{e}^{\underset{x\to \infty }{\mathop{lim}}\,\left( \frac{ax-4}{x} \right)}}\] \[={{e}^{2a}}={{e}^{3}}\] \[a=\frac{3}{2}\]
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