Question Number 80417 by jagoll last updated on 03/Feb/20
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} −{e}}{{x}}\:=\:? \\ $$
Commented by MJS last updated on 03/Feb/20
$$\mathrm{I}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{approximate}\:\mathrm{and}\:\mathrm{got}\:\mathrm{something} \\ $$$$\mathrm{very}\:\mathrm{close}\:\mathrm{to}\:−\frac{\mathrm{e}}{\mathrm{2}}\:\mathrm{but}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{prove}\:\mathrm{the} \\ $$$$\mathrm{exact}\:\mathrm{value} \\ $$
Commented by mr W last updated on 03/Feb/20
$$\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} ={e}\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{11}{x}^{\mathrm{2}} }{\mathrm{24}}−...\right) \\ $$$$\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} −{e}={e}\left(−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{11}{x}^{\mathrm{2}} }{\mathrm{24}}−...\right) \\ $$$$\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} −{e}}{{x}}={e}\left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{11}{x}}{\mathrm{24}}−...\right)\rightarrow−\frac{{e}}{\mathrm{2}} \\ $$
Commented by jagoll last updated on 03/Feb/20
$${via}\:{Taylor}\:{series}\:{sir}? \\ $$
Commented by mr W last updated on 03/Feb/20
$${yes}! \\ $$
Commented by jagoll last updated on 03/Feb/20
$${thank}\:{you}\:{sir} \\ $$