Question Number 211895 by Spillover last updated on 23/Sep/24
Answered by mehdee7396 last updated on 23/Sep/24
$${lim}_{{x}\rightarrow\infty} \:\left(\mathrm{1}+\frac{{a}}{{x}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)\mathrm{2}{x} \\ $$$$={lim}_{{x}\rightarrow\infty} \:\left(\frac{{ax}−\mathrm{4}}{{x}^{\mathrm{2}} }\right)\mathrm{2}{x}=\mathrm{2}{a} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\infty} \:\left(\mathrm{1}+\frac{{a}}{{x}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }−\mathrm{1}\right)^{\mathrm{2}{x}} ={e}^{\mathrm{2}{a}} \\ $$$$\Rightarrow\mathrm{2}{a}=\mathrm{3}\Rightarrow{a}=\frac{\mathrm{3}}{\mathrm{2}}\:\checkmark \\ $$$$ \\ $$
Answered by BHOOPENDRA last updated on 23/Sep/24
$${we}\:{know}\:{if}\:{we}\:{have}\:{function} \\ $$$$\underset{{x}\rightarrow{a}} {\mathrm{lim}}\:{f}\left({x}\right)^{{g}\left({x}\right)} =\mathrm{1}^{\infty} \\ $$$${then}\:{we}\:{use}\:{the}\:{property}\: \\ $$$$\underset{{e}^{{x}} \rightarrow{a}} {\mathrm{lim}}\:{g}\left({x}\right)\:\left\{{f}\left({x}\right)−\mathrm{1}\right\} \\ $$$${So}\:\underset{{e}^{{x}} \rightarrow\infty} {\mathrm{lim}}\:\mathrm{2}{x}\:\left(\mathrm{1}+\frac{{a}}{{x}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\:−\mathrm{1}\right)={e}^{\mathrm{3}} \\ $$$$\underset{{e}^{{x}} \rightarrow\infty} {\mathrm{lim}}\:\mathrm{2}{x}\:\left(\frac{{a}}{{x}}−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\right)={e}^{\mathrm{3}} \\ $$$$ \\ $$$${e}^{\mathrm{2}\left({a}−\frac{\mathrm{4}}{\infty}\right)} ={e}^{\mathrm{3}} \\ $$$${e}^{\mathrm{2}{a}} ={e}^{\mathrm{3}} \\ $$$$\mathrm{2}{a}=\mathrm{3} \\ $$$${a}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$