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Why does 0^0 scare people Watch

1. Why does it scare you??

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2. (Original post by Aph)
Why does it scare you??

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is that an emoticon?
3. (Original post by flou_fboco2)
is that an emoticon?
Nooo...

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4. (Original post by Aph)
Why does it scare you??

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5. it doenst scare me
6. (Original post by Aph)
Why does it scare you??

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What makes you think that it scares anyone?
7. Because some people don't like the mathematical answer.

I'm just shocked the OP put something they know scares people in the title, with no warning or anything
8. (Original post by TenOfThem)
What makes you think that it scares anyone?
Because many people deny its existence

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9. G-g-g-get it away

But seriously, I don't know

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10. (Original post by Aph)
Because many people deny its existence
Really

Are you sure they do not just believe that it is "undefined"
11. Doesnt scare me.

0^0=1
12. (Original post by TenOfThem)
Really

Are you sure they do not just believe that it is "undefined"
Well maybe but its obvious that it is equal to infinity

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13. (Original post by Aph)
Why does it scare you??

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Isn't it 1?

Haven't done Maths for ages...
14. (Original post by Aph)
Well maybe but its obvious that it is equal to infinity

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hmmmm
15. (Original post by Aph)
Well maybe but its obvious that it is equal to infinity

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We can easily show the limit of x^x as x approaches 0 is 1.

So why has it scared you?
16. (Original post by newblood)
Doesnt scare me nor anyone. Simple limits.

0^0=1
Not as simple as you think. If exists then . This is obviously not true for all paths to in .
17. (Original post by newblood)
We can easily show the limit of x^x as x approaches 0 is 1.

So why has it scared you?
Ok but surely it equals all numbers
It hasn't though the weird things thst happen to x^y=y^x around e does. Though its also quite beautiful so id love an explanation

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18. *Runs away from thread in terror*
19. (Original post by FireGarden)
Not as simple as you think. If exists then . This is obviously not true for all paths to in .
Take x^x=exp(xlnx). It suffices to show lim as x approaches 0 of xlnx is 0 which is easily done using l'hopitals by noting that x=1/(1/x)
20. (Original post by newblood)
Take x^x=exp(xlnx). It suffices to show lim as x approaches 0 of xlnx is 0 which is easily done using l'hopitals by noting that x=1/(1/x)
This limit is not good enough. You can of course have a number to the power of a different number, and to take that into account, you need to approach from any possibility; therefore we must consider . Now I agree, , but that is only one case of the required limit (along the line y=x towards the origin).

Consider the limits ; and . These limits are clearly not equal, hence there is no unique, sensible value for .

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