Question Number 80690 by ahmadshahhimat775@gmail.com last updated on 05/Feb/20
Commented by jagoll last updated on 05/Feb/20
$$\underset{{x}\rightarrow{e}} {\mathrm{lim}}\:\frac{{xlnx}−{x}}{\mathrm{2}−\mathrm{2}{lnx}}\:=\underset{{x}\rightarrow{e}} {\mathrm{lim}}\:\frac{{x}\left({lnx}\:−\mathrm{1}\right)}{−\mathrm{2}\left({lnx}−\mathrm{1}\right)}= \\ $$$$−\frac{{e}}{\mathrm{2}} \\ $$
Commented by abdomathmax last updated on 05/Feb/20
$${let}\:{f}\left({x}\right)=\frac{{ln}\left({x}^{{x}} \right)−{x}}{\mathrm{2}−{ln}\left({x}^{\mathrm{2}} \right)}\:\Rightarrow{f}\left({x}\right)=\frac{{xln}\left({x}\right)−{x}}{\mathrm{2}−\mathrm{2}{lnx}} \\ $$$${let}\:{x}={e}+{u}\:\Rightarrow{f}\left({x}\right)={g}\left({u}\right)=\frac{\left({e}+{u}\right){ln}\left({e}+{u}\right)−{u}−{e}}{\mathrm{2}\left\{\mathrm{1}−{ln}\left({e}+{u}\right)\right\}} \\ $$$${x}\rightarrow{e}\:\Rightarrow{u}\rightarrow\mathrm{0}\:\Rightarrow \\ $$$${g}\left({u}\right)=\frac{\left({e}+{u}\right)\left(\mathrm{1}+{ln}\left(\mathrm{1}+\frac{{u}}{{e}}\right)\right)−{u}−{e}}{\mathrm{2}\left\{\mathrm{1}−{ln}\left({e}\right)−{ln}\left(\mathrm{1}+\frac{{u}}{{e}}\right)\right\}} \\ $$$$=\frac{\left({e}+{u}\right){ln}\left(\mathrm{1}+\frac{{u}}{{e}}\right)}{−\mathrm{2}{ln}\left(\mathrm{1}+\frac{{u}}{{e}}\right)}\:\Rightarrow{g}\left({u}\right)\sim\:\frac{\left({e}+{u}\right)\frac{{u}}{{e}}}{−\mathrm{2}×\frac{{u}}{{e}}}\:\:\:\left({u}\:\in{V}\left(\mathrm{0}\right)\right) \\ $$$$\Rightarrow{g}\left({u}\right)\sim−\:\frac{{u}+\frac{{u}^{\mathrm{2}} }{{e}}}{\mathrm{2}}×\frac{{e}}{{u}}=−\frac{{e}}{\mathrm{2}}\left(\mathrm{1}+\frac{{u}}{{e}}\right)\rightarrow−\frac{{e}}{\mathrm{2}}\:\left({u}\rightarrow\mathrm{0}\right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow{e}} \:\:\:{f}\left({x}\right)=−\frac{{e}}{\mathrm{2}} \\ $$