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Question Number 163931 by mathlove last updated on 12/Jan/22

$$\underset{x\to 0}{\lim }\frac{x^{x^x}}{x}=?$$$$pleashelp$$

Answered by mahdipoor last updated on 12/Jan/22

$$\underset{x\to 0}{\lim }\frac{lnx}{\frac{1}{x^x-1}}=\frac{-\mathrm{\infty }}{\mathrm{\infty }}\stackrel{hop}{\Rightarrow }\underset{x\to 0}{\lim }\frac{\frac{1}{x}}{\frac{-x^x(xlnx+1)}{{(x^x-1)}^2}}=$$$$\frac{\underset{x\to 0}{\lim }\frac{{(x^x-1)}^2}{x}}{\underset{x\to 0}{\lim }-x^x(xlnx+1)}=\frac{\underset{x\to 0}{\lim }\frac{2(x^x-1)x^x(xlnx+1)}{1}}{-1(0+1)}=$$$$\frac{\frac{2\times 0\times 1\times (0+1)}{1}}{-1}=0$$$$\Rightarrow \underset{x\to 0}{\lim }\frac{x^{x^x}}{x}=\underset{x\to 0}{\lim }{x}^{(x^x-1)}=e^{\underset{x\to 0}{\lim }\frac{lnx}{1/(x^x-1)}}=e^0=1$$

Commented by mathlove last updated on 12/Jan/22

$$thanks$$

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