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Question Number 30256 by mondodotto@gmail.com last updated on 19/Feb/18

Commented by abdo imad last updated on 19/Feb/18

in this case is better to use hospital theorem let put  u(x)=x^x  −x and v(x)=1−x +lnx wehave u(1)=v(1)=0  u(x)=e^(xlnx) −x⇒u^′ (x)=(1+lnx)e^(xlnx)  −1 and  u^(′′) (x)=(1/x) e^(xlnx)  +(1+lnx)(1+lnx)e^(xlnx)   =(1/x) +(1+lnx)^2  e^(xlnx)   ⇒ u^(′′) (1)= 2  v^′ (x)= −1 +(1/x) ⇒ v^(′′) (x)=−(1/x^2 ) ⇒v^(′′) (1)=−1 so  lim_(x→1)   ((x^x  −x)/(1−x +lnx)) =−2 .

$${in}\:{this}\:{case}\:{is}\:{better}\:{to}\:{use}\:{hospital}\:{theorem}\:{let}\:{put} \\ $$$${u}\left({x}\right)={x}^{{x}} \:−{x}\:{and}\:{v}\left({x}\right)=\mathrm{1}−{x}\:+{lnx}\:{wehave}\:{u}\left(\mathrm{1}\right)={v}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$${u}\left({x}\right)={e}^{{xlnx}} −{x}\Rightarrow{u}^{'} \left({x}\right)=\left(\mathrm{1}+{lnx}\right){e}^{{xlnx}} \:−\mathrm{1}\:{and} \\ $$$${u}^{''} \left({x}\right)=\frac{\mathrm{1}}{{x}}\:{e}^{{xlnx}} \:+\left(\mathrm{1}+{lnx}\right)\left(\mathrm{1}+{lnx}\right){e}^{{xlnx}} \\ $$$$=\frac{\mathrm{1}}{{x}}\:+\left(\mathrm{1}+{lnx}\right)^{\mathrm{2}} \:{e}^{{xlnx}} \:\:\Rightarrow\:{u}^{''} \left(\mathrm{1}\right)=\:\mathrm{2} \\ $$$${v}^{'} \left({x}\right)=\:−\mathrm{1}\:+\frac{\mathrm{1}}{{x}}\:\Rightarrow\:{v}^{''} \left({x}\right)=−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow{v}^{''} \left(\mathrm{1}\right)=−\mathrm{1}\:{so} \\ $$$${lim}_{{x}\rightarrow\mathrm{1}} \:\:\frac{{x}^{{x}} \:−{x}}{\mathrm{1}−{x}\:+{lnx}}\:=−\mathrm{2}\:. \\ $$

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