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lim-x-0-1-x-1-x-e-x-




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

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