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




Question Number 163657 by mathlove last updated on 09/Jan/22
lim_(x→0) (((1+x)^(1/x) −e)/x)=?
$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} −{e}}{{x}}=? \\ $$
Answered by mr W last updated on 09/Jan/22
=lim_(x→0) (e^((ln (1+x))/x) )′  =lim_(x→0) (1+x)^(1/x) ((1/(1+x))−(1/x)ln (1+x))(1/x)  =lim_(x→0) (1+x)^(1/x) (1−x+x^2 −...−(((x−(x^2 /2)+(x^3 /3)−...))/x))(1/x)  =lim_(x→0) (1+x)^(1/x) (1−x+x^2 −...−(1−(x/2)+(x^2 /3)−...))(1/x)  =lim_(x→0) (1+x)^(1/x) (−(x/2)+((2x^2 )/3)−....)(1/x)  =lim_(x→0) (1+x)^(1/x) (−(1/2)+((2x)/3)−....)  =e(−(1/2))  =−(e/2)
$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left({e}^{\frac{\mathrm{ln}\:\left(\mathrm{1}+{x}\right)}{{x}}} \right)' \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \left(\frac{\mathrm{1}}{\mathrm{1}+{x}}−\frac{\mathrm{1}}{{x}}\mathrm{ln}\:\left(\mathrm{1}+{x}\right)\right)\frac{\mathrm{1}}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \left(\mathrm{1}−{x}+{x}^{\mathrm{2}} −…−\frac{\left({x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}−…\right)}{{x}}\right)\frac{\mathrm{1}}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \left(\mathrm{1}−{x}+{x}^{\mathrm{2}} −…−\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}−…\right)\right)\frac{\mathrm{1}}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \left(−\frac{{x}}{\mathrm{2}}+\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}}−….\right)\frac{\mathrm{1}}{{x}} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{1}+{x}\right)^{\frac{\mathrm{1}}{{x}}} \left(−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{2}{x}}{\mathrm{3}}−….\right) \\ $$$$={e}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$=−\frac{{e}}{\mathrm{2}} \\ $$
Commented by mathlove last updated on 09/Jan/22
thanks  mr  W
$${thanks}\:\:{mr}\:\:{W} \\ $$
Answered by mnjuly1970 last updated on 09/Jan/22
   lim_( x→0)  (( e^( (1/x) ln(1+x)) −e)/x)    =lim_( x→0) (( e^( (1/x) ( x−(x^( 2) /2) + O(x^( 2) ))) −e)/x)   = elim_(  x→0) (( −(1/2) e^( −x) )/1) =((−e)/2)
$$\:\:\:{lim}_{\:{x}\rightarrow\mathrm{0}} \:\frac{\:{e}^{\:\frac{\mathrm{1}}{{x}}\:{ln}\left(\mathrm{1}+{x}\right)} −{e}}{{x}} \\ $$$$\:\:={lim}_{\:{x}\rightarrow\mathrm{0}} \frac{\:{e}^{\:\frac{\mathrm{1}}{{x}}\:\left(\:{x}−\frac{{x}^{\:\mathrm{2}} }{\mathrm{2}}\:+\:{O}\left({x}^{\:\mathrm{2}} \right)\right)} −{e}}{{x}} \\ $$$$\:=\:{elim}_{\:\:{x}\rightarrow\mathrm{0}} \frac{\:−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{\:−{x}} }{\mathrm{1}}\:=\frac{−{e}}{\mathrm{2}} \\ $$
Commented by mathlove last updated on 10/Jan/22
thanks
$${thanks} \\ $$

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