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Can you prove that \(0^0=1\) ?

Note by Bruce Wayne 3 years, 11 months ago

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Hi this question has been asked repeatedly on brilliant,

\(0^0 = e^{0 \ln 0}\) is undefined , but we can find \(\displaystyle \lim_{x \to 0} x^x = \lim_{x \to 0} e^{x \ln x } = 1\) , as \(\displaystyle \lim_{x \to 0} x\ln x = 0 \) ,

This is very similar to saying that \(\frac{0}{0} \) is undefined but \(\displaystyle\lim_{x \to 0 } \frac{x}{x}\) is defined.

Also \(\displaystyle\lim_{(f(x) \to 0, g(x) \to 0} {\big(f(x)\big)}^{g(x)}\) is not \(1\) , it would depend on \(f(x)\) and \(g(x)\)

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Maybe this can help.

0^0 is undefined...

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TopNewestHi this question has been asked repeatedly on brilliant,

\(0^0 = e^{0 \ln 0}\) is undefined , but we can find \(\displaystyle \lim_{x \to 0} x^x = \lim_{x \to 0} e^{x \ln x } = 1\) , as \(\displaystyle \lim_{x \to 0} x\ln x = 0 \) ,

This is very similar to saying that \(\frac{0}{0} \) is undefined but \(\displaystyle\lim_{x \to 0 } \frac{x}{x}\) is defined.

Also \(\displaystyle\lim_{(f(x) \to 0, g(x) \to 0} {\big(f(x)\big)}^{g(x)}\) is not \(1\) , it would depend on \(f(x)\) and \(g(x)\)

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Maybe this can help.

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0^0 is undefined...

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