Question Number 194685 by cortano12 last updated on 13/Jul/23
Answered by qaz last updated on 14/Jul/23
$$\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{1}}{{x}}{ln}\left(\mathrm{1}+{x}\right)} ={e}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…} \\ $$$$={e}\left(\mathrm{1}+\left(−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{2}} +…\right)^{\mathrm{2}} +…\right. \\ $$$$={e}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{11}}{\mathrm{24}}{x}^{\mathrm{2}} +{o}\left({x}^{\mathrm{2}} \right)\right) \\ $$$$\Rightarrow\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} −{e}+\frac{\mathrm{1}}{\mathrm{2}}{ex}}{{x}^{\mathrm{2}} }=\frac{\mathrm{11}}{\mathrm{24}}{e} \\ $$