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Question Number 88610 by M±th+et£s last updated on 11/Apr/20

$$is1\times 1\times 1\times ..........=1^{\mathrm{\infty }}$$$$or$$$$1\times 1\times 1\times 1\times 1\times ..........=1$$$$$$$$$$

Commented by mahdi last updated on 11/Apr/20

$$\mathrm{for}\mathrm{example}:$$$$\mathrm{A})\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{\mathrm{x}})}^{\mathrm{x}}=1^{\mathrm{\infty }}=\mathrm{e}\ne 1$$$$\mathrm{B})\underset{x\to \mathrm{\infty }}{\lim }\left(1^{\mathrm{x}}\right)=1^{\mathrm{\infty }}=1=1$$$$\mathrm{in}\mathrm{A}:1\times 1\times ...=1^{\mathrm{\infty }}\ne 1\mathrm{B}:1\times 1\times ...=1^{\mathrm{\infty }}=1$$$$\mathrm{Question}\mathrm{is}\mathrm{important}.$$

Commented by M±th+et£s last updated on 11/Apr/20

$$andthereismoreexampleslike$$$$\underset{x\to \mathrm{\infty }}{lim}{(1+\frac{1}{x^2})}^x=1$$$$\underset{x\to \mathrm{\infty }}{lim}(1+\frac{1}{log\left(x\right)})=\mathrm{\infty }$$$$\underset{x\to \mathrm{\infty }}{lim}{(1-\frac{1}{log\left(x\right)})}^x=0$$$$$$$$soithinkthatitsnotwrongifwesaythat1\times 1\times 1\times 1\times .....=1^{\mathrm{\infty }}=1$$$$butnotineverytime1\times 1\times 1\times 1.....=1^{\mathrm{\infty }}$$$$$$$$$$

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