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Question Number 36843 by Penguin last updated on 06/Jun/18

$$\underset{x\to \mathrm{\infty }}{\lim }\left(x^{x^{x^{.....}}}\right)$$

Commented by maxmathsup by imad last updated on 06/Jun/18

$$letput{A}_n\left(x\right)=x^{x^{x...}}\left(n\times \right)$$$$A_2\left(x\right)=x^x=e^{xln\left(x\right)}$$$$A_3\left(x\right)=x^{x^x}={\left(A_2\left(x\right)\right)}^x=e^{x^{2}ln\left(x\right)}\Rightarrow A_n\left(x\right)=e^{x^{n-1}ln\left(x\right)}letsupposethat$$$$A_{n+1}\left(x\right)=x^{x^{...}}(n+1)\times ={\left(A_n\left(x\right)\right)}^x={\left(e^{x^{n-1}ln\left(x\right)}\right)}^x=e^{x^{n}ln\left(x\right)}$$$$wehave{lim}_{x\to +\mathrm{\infty }andn\to +\mathrm{\infty }}{A}_n\left(x\right)=+\mathrm{\infty }.$$

Answered by MJS last updated on 06/Jun/18

$$\mathrm{\infty },\mathrm{what}\mathrm{else}\mathrm{would}\mathrm{you}\mathrm{expect}?$$

Commented by maxmathsup by imad last updated on 06/Jun/18

$$thatmustbe+\mathrm{\infty }$$

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