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Question Number 96319 by  M±th+et+s last updated on 31/May/20
lim_(x→0) ((((1+x)^(1/x) )/e))^(1/x)
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} }{{e}}\right)^{\frac{\mathrm{1}}{{x}}} \\ $$
Answered by abdomathmax last updated on 31/May/20
f(x)=((((1+x)^(1/x) )/e))^(1/x) ⇒ln(f(x))=(1/x)ln((((1+x)^(1/x) )/e)) =(1/x)( ln(1+x)^(1/x) −1) =(1/x)((1/x)ln(1+x)−1) =((ln(1+x)−x)/x^2 ) hodpiral →lim_(x→0) ((ln(1+x)−x)/x^2 ) =lim_(x→0) (((1/(1+x))−1)/(2x)) =lim_(x→0) ((−1)/(2(1+x^2 ))) =−(1/2) ⇒ ln(f(x))→−(1/2) ⇒f(x)→e^(−(1/2)) =(1/( (√e)))
$$\mathrm{f}\left(\mathrm{x}\right)=\left(\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{e}}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \:\Rightarrow\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} }{\mathrm{e}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}}\left(\:\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}}\left(\frac{\mathrm{1}}{\mathrm{x}}\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)−\mathrm{1}\right)\:=\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)−\mathrm{x}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{hodpiral}\:\rightarrow\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)−\mathrm{x}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\:\:\frac{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}−\mathrm{1}}{\mathrm{2x}}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\:\:\:\frac{−\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{ln}\left(\mathrm{f}\left(\mathrm{x}\right)\right)\rightarrow−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\rightarrow\mathrm{e}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{e}}} \\ $$
Commented by  M±th+et+s last updated on 31/May/20
nice work sir thank you
$${nice}\:{work}\:{sir}\:{thank}\:{you} \\ $$
Commented by mathmax by abdo last updated on 31/May/20
you are welcome sir.
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome}\:\mathrm{sir}. \\ $$

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