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    Why does it scare you??

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

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

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

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    :afraid:

    :dontknow:

    :iiam:
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    :holmes: it doenst scare me
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    (Original post by Aph)
    Why does it scare you??

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    What makes you think that it scares anyone?
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    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
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    (Original post by TenOfThem)
    What makes you think that it scares anyone?
    Because many people deny its existence

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    G-g-g-get it away :afraid:

    But seriously, I don't know

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

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

    0^0=1
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    (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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    (Original post by Aph)
    Why does it scare you??

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    Isn't it 1?

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

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    hmmmm
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    (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?
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    (Original post by newblood)
    Doesnt scare me nor anyone. Simple limits.

    0^0=1
    Not as simple as you think. If 0^0 exists then \lim_{x\to 0}(\lim_{y\to 0}(x^y)) = \lim_{y\to 0}(\lim_{x\to 0}(x^y)) . This is obviously not true for all paths to (0,0) in \mathbb{R}^2.
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    (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 :confused:
    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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    *Runs away from thread in terror* :eek:
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    (Original post by FireGarden)
    Not as simple as you think. If 0^0 exists then \lim_{x\to 0}(\lim_{y\to 0}(x^y)) = \lim_{y\to 0}(\lim_{x\to 0}(x^y)) . This is obviously not true for all paths to (0,0) in \mathbb{R}^2.
    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)
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    (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 0^0 from any possibility; therefore we must consider x^y. Now I agree, \lim_{x\to 0}(x^x)=1, but that is only one case of the required limit (along the line y=x towards the origin).

    Consider the limits \lim_{x\to 0}(x^0)=1; and \lim_{x\to 0}(0^x)=0. These limits are clearly not equal, hence there is no unique, sensible value for 0^0.
 
 
 
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