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Question Number 75701 by vishalbhardwaj last updated on 15/Dec/19

$$\mathrm{please}\mathrm{tell}\mathrm{me}:{\left(\mathrm{\infty }\right)}^0\mathrm{is}\mathrm{not}\mathrm{defined}?$$

Commented by malwaan last updated on 16/Dec/19

$$yes$$$$but{0}^{\mathrm{\infty }}=0$$

Answered by MJS last updated on 15/Dec/19

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{defined}$$

Answered by aguy last updated on 15/Dec/19

$${\mathrm{It}}^{,}\mathrm{s}\mathrm{not}\mathrm{defined}\mathrm{because}\mathrm{infinite}\mathrm{is}\mathrm{not}$$$$\mathrm{a}\mathrm{number},\mathrm{but}{lim}_{x\to \mathrm{\infty }}\left(x^0\right)=1.\mathrm{It}$$$$\mathrm{happens}\mathrm{because}\mathrm{if}x\mathrm{is}\mathrm{a}\mathrm{positive}\mathrm{real}$$$$\mathrm{number},\mathrm{then}{x}^0=1.$$

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