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Can you prove that $0^0=1$ ?

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Maybe this can help.

Vincent Tandya - 7 years, 6 months ago

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 $g(x)$

jatin yadav - 7 years, 6 months ago

0^0 is undefined...

John Ashley Capellan - 7 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$