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Question Number 203909 by Davidtim last updated on 01/Feb/24

$$provethat{0}^n=0,n>0$$

Answered by Mathspace last updated on 01/Feb/24

$$0^n={lim}_{x\to 0^{+}}{x}^n$$$$={lim}_{x\to 0^{+}}{e}^{nlogx}but$$$${lim}_{x\to 0^{+}}nlogx=-\mathrm{\infty }(n>0)$$$$\Rightarrow 0^n=e^{-\mathrm{\infty }}=0$$

Answered by witcher3 last updated on 02/Feb/24

$$0^{\mathrm{n}}=0;\mathrm{n}=1$$$$0^{\mathrm{n}}=0.0^{\mathrm{n}-1}=0;\mathrm{\forall }\mathrm{n}⩾2$$$$\mathrm{for}\mathrm{n}⩾2;\mathrm{n}-1⩾1\Rightarrow 0^{\mathrm{n}-1}\mathrm{is}\mathrm{defined}$$$$0.0^{\mathrm{n}-1}=0$$$$\Rightarrow \mathrm{\forall }\mathrm{n}⩾10^{\mathrm{n}}=0$$$$$$

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