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Question Number 160625 by Jamshidbek last updated on 03/Dec/21

$$\underset{x\to 0+0}{\lim }({\mathrm{x}}^{\mathrm{x}}-1)\mathrm{lnx}$$

Answered by Mathspace last updated on 03/Dec/21

$$f\left(x\right)=(x^x-1)lnx\Rightarrow$$$$f\left(x\right){=}_{lnx=t}{\left(e^t\right)}^{e^t}-1)t=(e^{{te}^t}-1)t$$$$(t\to -\mathrm{\infty })$$$$g\left(t\right)=t(e^{{te}^t}-1)=t{e}^{{te}^t}-t$$$${lim}_{t\to -\mathrm{\infty }}{te}^{{te}^t}=0because{e}^{t}defeat{t}^n\Rightarrow$$$${lim}_{t\to -\mathrm{\infty }}g\left(t\right)=0={lim}_{x\to 0^{+}}f\left(x\right)$$

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