Question Number 209810 by mnjuly1970 last updated on 22/Jul/24
$$ \\ $$$$\:\:\:\mathrm{lim}_{\:\mathrm{x}\rightarrow\mathrm{0}} \:\frac{\:\left(\mathrm{1}+\:\mathrm{x}\:\right)^{\frac{\mathrm{1}}{\mathrm{x}}} −\mathrm{e}}{\mathrm{x}}\:=\:? \\ $$$$ \\ $$
Answered by mr W last updated on 22/Jul/24
![(1+x)^(1/x) =e^((ln (1+x))/x) =e^(1−(x/2)+(x^2 /3)−(x^3 /4)+...) [(1+x)^(1/x) ]^, =e^(1−(x/2)+(x^2 /3)−(x^3 /4)+...) (−(1/2)+((2x)/3)−((3x^2 )/4)+...) L=(([(1+x)^(1/x) ]^, −0)/1)=[(1+x)^(1/x) ]_(x=0) ^, =e(−(1/2))=−(e/2) ✓](https://www.tinkutara.com/question/Q209819.png)
$$\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}} ={e}^{\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{{x}^{\mathrm{3}} }{\mathrm{4}}+…} \\ $$$$\left[\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \right]^{,} ={e}^{\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−\frac{{x}^{\mathrm{3}} }{\mathrm{4}}+…} \left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}{x}}{\mathrm{3}}−\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}}+…\right) \\ $$$${L}=\frac{\left[\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \right]^{,} −\mathrm{0}}{\mathrm{1}}=\left[\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \right]_{{x}=\mathrm{0}} ^{,} \\ $$$$\:\:\:={e}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=−\frac{{e}}{\mathrm{2}}\:\checkmark \\ $$
Commented by mnjuly1970 last updated on 22/Jul/24
$${thanks}\:{alot}\:{sir} \\ $$$$ \\ $$