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Question Number 64011 by Prithwish sen last updated on 12/Jul/19

lim_(x→0) ((x^x −1)/(xlnx))

$$\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\frac{\mathrm{x}^{\mathrm{x}} −\mathrm{1}}{\mathrm{xlnx}} \\ $$

Commented by kaivan.ahmadi last updated on 12/Jul/19

y=x^x ⇒lny=xlnx⇒((y′)/y)=lnx+x×(1/x)=1+lnx⇒  y′=y(1+lnx)=x^x (1+lnx)  lim_(x→0^+ ) ((x^x −1)/(xlnx))=lim_(x→0^+ ) ((x^x −1)/(lnx^x ))  =^(hop)   lim_(x→0^+ )   ((x^x (1+lnx))/((x^x (1+lnx))/x^x ))=lim_(x→0^+ ) x^x =  lim_(x→0^+ ) e^(xlnx) =e^0 =1

$${y}={x}^{{x}} \Rightarrow{lny}={xlnx}\Rightarrow\frac{{y}'}{{y}}={lnx}+{x}×\frac{\mathrm{1}}{{x}}=\mathrm{1}+{lnx}\Rightarrow \\ $$$${y}'={y}\left(\mathrm{1}+{lnx}\right)={x}^{{x}} \left(\mathrm{1}+{lnx}\right) \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \frac{{x}^{{x}} −\mathrm{1}}{{xlnx}}={lim}_{{x}\rightarrow\mathrm{0}^{+} } \frac{{x}^{{x}} −\mathrm{1}}{{lnx}^{{x}} }\:\:\overset{{hop}} {=} \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\frac{{x}^{{x}} \left(\mathrm{1}+{lnx}\right)}{\frac{{x}^{{x}} \left(\mathrm{1}+{lnx}\right)}{{x}^{{x}} }}={lim}_{{x}\rightarrow\mathrm{0}^{+} } {x}^{{x}} = \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } {e}^{{xlnx}} ={e}^{\mathrm{0}} =\mathrm{1} \\ $$

Commented by mathmax by abdo last updated on 12/Jul/19

let A(x)=((x^x −1)/(xlnx)) ⇒A(x)=((e^(xln(x)) −1)/(xlnx))   we have lim_(x→0^+ )   xln(x)=0  and e^u  ∼ 1+u   (u∈V(0)) ⇒e^(xln(x)) =1+xln(x) ⇒  e^(xln(x)) −1 ∼xln(x)  (x→0) ⇒((e^(xln(x)) −1)/(xlnx)) ∼1 ⇒  lim_(x→0^+ )    A(x) =1 .

$${let}\:{A}\left({x}\right)=\frac{{x}^{{x}} −\mathrm{1}}{{xlnx}}\:\Rightarrow{A}\left({x}\right)=\frac{{e}^{{xln}\left({x}\right)} −\mathrm{1}}{{xlnx}}\:\:\:{we}\:{have}\:{lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:{xln}\left({x}\right)=\mathrm{0} \\ $$$${and}\:{e}^{{u}} \:\sim\:\mathrm{1}+{u}\:\:\:\left({u}\in{V}\left(\mathrm{0}\right)\right)\:\Rightarrow{e}^{{xln}\left({x}\right)} =\mathrm{1}+{xln}\left({x}\right)\:\Rightarrow \\ $$$${e}^{{xln}\left({x}\right)} −\mathrm{1}\:\sim{xln}\left({x}\right)\:\:\left({x}\rightarrow\mathrm{0}\right)\:\Rightarrow\frac{{e}^{{xln}\left({x}\right)} −\mathrm{1}}{{xlnx}}\:\sim\mathrm{1}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}^{+} } \:\:\:{A}\left({x}\right)\:=\mathrm{1}\:. \\ $$

Commented by Prithwish sen last updated on 12/Jul/19

thank you sirs.

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sirs}. \\ $$

Answered by Rio Michael last updated on 12/Jul/19

(lnx+1)x^x   let y= x^x    since ((d(−1))/dx)=0  lny = lnx^x    (1/y)= x((1/x))+ lnx    (1/y)= lnx + 1      = (lnx+1)x^x   lim_(x→0) (((lnx+1)x^x )/(lnx+1))   lim_(x→0)  x^x   = 0     please check

$$\left({lnx}+\mathrm{1}\right){x}^{{x}} \\ $$$${let}\:{y}=\:{x}^{{x}} \:\:\:{since}\:\frac{{d}\left(−\mathrm{1}\right)}{{dx}}=\mathrm{0} \\ $$$${lny}\:=\:{lnx}^{{x}} \\ $$$$\:\frac{\mathrm{1}}{{y}}=\:{x}\left(\frac{\mathrm{1}}{{x}}\right)+\:{lnx} \\ $$$$\:\:\frac{\mathrm{1}}{{y}}=\:{lnx}\:+\:\mathrm{1} \\ $$$$\:\:\:\:=\:\left({lnx}+\mathrm{1}\right){x}^{{x}} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\frac{\left({lnx}+\mathrm{1}\right){x}^{{x}} }{{lnx}+\mathrm{1}} \\ $$$$\:\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{x}^{{x}} \\ $$$$=\:\mathrm{0}\:\:\: \\ $$$${please}\:{check}\: \\ $$

Commented by Prithwish sen last updated on 12/Jul/19

thamk you sir

$$\mathrm{thamk}\:\mathrm{you}\:\mathrm{sir} \\ $$

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