lim-x-0-x-x-x-

Question Number 60006 by meme last updated on 17/May/19

$${li}\underset{{x}\rightarrow\mathrm{0}} {{m}}\frac{{x}^{{x}} }{{x}}=? \\ $$

Commented by maxmathsup by imad last updated on 17/May/19

let A(x) =(x^x /x) for x>0 ⇒A(x) =(e^(xln(x)) /x) we have lim_(x→0^+ ) e^(xln(x)) =1 but e^u ∼1 +u ⇒e^(xlnx) ∼1+xln(x) ⇒(e^(xln(x)) /x)∼(1/x) +ln(x) =((1+xln(x))/x) →+∞(x→0^+ ) ⇒lim_(x→0^+ ) A(x) =+∞ another way we have A(x)=x^(x−1) =e^((x−1)ln(x)) =e^(xln(x)−ln(x)) →e^(+∞) =+∞ (x→0^+ )

$${let}\:{A}\left({x}\right)\:=\frac{{x}^{{x}} }{{x}}\:\:\:{for}\:{x}>\mathrm{0}\:\Rightarrow{A}\left({x}\right)\:=\frac{{e}^{{xln}\left({x}\right)} }{{x}}\:\:\:\:{we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{e}^{{xln}\left({x}\right)} =\mathrm{1}\:\:{but} \\ $$$${e}^{{u}} \:\sim\mathrm{1}\:\:+{u}\:\:\Rightarrow{e}^{{xlnx}} \sim\mathrm{1}+{xln}\left({x}\right)\:\Rightarrow\frac{{e}^{{xln}\left({x}\right)} }{{x}}\sim\frac{\mathrm{1}}{{x}}\:+{ln}\left({x}\right)\:=\frac{\mathrm{1}+{xln}\left({x}\right)}{{x}}\:\rightarrow+\infty\left({x}\rightarrow\mathrm{0}^{+} \right) \\ $$$$\Rightarrow{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{A}\left({x}\right)\:=+\infty \\ $$$${another}\:{way}\:\:{we}\:{have}\:{A}\left({x}\right)={x}^{{x}−\mathrm{1}} \:={e}^{\left({x}−\mathrm{1}\right){ln}\left({x}\right)} \:={e}^{{xln}\left({x}\right)−{ln}\left({x}\right)} \:\rightarrow{e}^{+\infty} =+\infty \\ $$$$\left({x}\rightarrow\mathrm{0}^{+} \right) \\ $$

Commented by kaivan.ahmadi last updated on 19/May/19