Question Number 70364 by 20190927 last updated on 03/Oct/19
$$\mathrm{solve}\:\mathrm{L}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{e}^{\mathrm{x}} −\frac{\mathrm{1}}{\mathrm{1}−\mathrm{x}}}{\mathrm{x}^{\mathrm{2}} } \\ $$
Commented by kaivan.ahmadi last updated on 03/Oct/19
$${lim}_{{x}\rightarrow\mathrm{0}} \frac{\left(\mathrm{1}−{x}\right){e}^{{x}} −\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}={lim}_{{x}\rightarrow\mathrm{0}} \frac{−{xe}^{{x}} }{\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{2}} }= \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \frac{−{e}^{{x}} −{xe}^{{x}} }{\mathrm{2}−\mathrm{6}{x}}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by 20190927 last updated on 03/Oct/19
$$\mathrm{Thank}\:\mathrm{you}\: \\ $$
Commented by mathmax by abdo last updated on 03/Oct/19
$${let}\:{f}\left({x}\right)=\frac{{e}^{{x}} −\frac{\mathrm{1}}{\mathrm{1}−{x}}}{{x}^{\mathrm{2}} }\:\Rightarrow{f}\left({x}\right)\:=\frac{\left(\mathrm{1}−{x}\right){e}^{{x}} −\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)} \\ $$$${we}\:{have}\:{e}^{{x}} =\mathrm{1}+{x}\:+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{3}} \right)\:\Rightarrow\left(\mathrm{1}−{x}\right){e}^{{x}} \:=\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{3}} \right)\right) \\ $$$$=\mathrm{1}+{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−{x}−{x}^{\mathrm{2}} −\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{4}} \right)=\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{2}}\:+{o}\left({x}^{\mathrm{4}} \right)\:\Rightarrow \\ $$$${f}\left({x}\right)\sim\frac{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{{x}^{\mathrm{3}} }{\mathrm{2}}}{{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}\:\Rightarrow{f}\left({x}\right)\:\sim\:\:\frac{−\frac{\mathrm{1}}{\mathrm{2}}−\frac{{x}}{\mathrm{2}}}{\mathrm{1}−{x}}\:\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:\:{f}\left({x}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$