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