In fairness, when considering 0^0, it's kind of important to understand why x^0 is 1 (@1:45) and 0^x is 0. As for the answer to 0^0, using the limit with real numbers basically definitively denotes 0^0 as 1 (a convergent limit isn't really an "it's close, but I don't know" scenario). The reason 0^0 is considered undefined rather than equal to 1 is, to my understanding, due to indeterminate limits when utilizing complex numbers. So similarly to how 1/x has separate limits for x=0 depending on if approaching from left or right, 0^0 has limits other than 1 is approaching from complex directions.
0^0 comes to a problem of being undefined due to the arithmetic-geometric transfer. In the arithmetic-geometric transfer, all 0's become 1's and the operation hierarchy moves up by 1 level: Addition to multiplication, multiplication to exponentation, and exponentation to tetration. Therefore, 0^0=1^^1, or 1 tetrated to 1. However, 1 tetrated to 1 is just a power tower of one 1, so if 0^0 is undefined, 1 tetrated to 1 is also indeterminate, and all real and complex numbers are UNDEFINED! Also, 0 factorial has the same discrepancy as we know that factorials of nonnegative numbers is the product of all numbers less than or equal to the number. Since 0 is an annihilator, 0 isn't included in the product. Therefore, 1 is the lowest number that can use the factorial definition. However, we know that (-1)! is infinity, so 0!=(-1!)*0 (due to the factorial rules), but that is infinity*0, which is also undefined. Therefore, 0^0 and 0! live and die together: If 0! is 1, 0^0 is also 1, and if 0^0 is undefined, 0! is also undefined.
0^0 is undefined. According to your situation by using limits, the limit just comes out to be 1. This is according to the graph of y=x^x. Notice that when x=0, that value is undefined and has a point of discontinuity. The limit comes from the use of l'Hopital's rule if you want to know how to get the limit to turn to be 1.
0^0 comes to a problem of being undefined due to the arithmetic-geometric transfer. In the arithmetic-geometric transfer, all 0's become 1's and the operation hierarchy moves up by 1 level: Addition to multiplication, multiplication to exponentation, and exponentation to tetration. Therefore, 0^0=1^^1, or 1 tetrated to 1. However, 1 tetrated to 1 is just a power tower of one 1, so if 0^0 is undefined, 1 tetrated to 1 is also indeterminate, and all real and complex numbers are UNDEFINED! Also, 0 factorial has the same discrepancy as we know that factorials of nonnegative numbers is the product of all numbers less than or equal to the number. Since 0 is an annihilator, 0 isn't included in the product. Therefore, 1 is the lowest number that can use the factorial definition. However, we know that (-1)! is infinity, so 0!=(-1!)*0 (due to the factorial rules), but that is infinity*0, which is also undefined. Therefore, 0^0 and 0! live and die together: If 0! is 1, 0^0 is also 1, and if 0^0 is undefined, 0! is also undefined.
0^0 comes to a problem of being undefined due to the arithmetic-geometric transfer. In the arithmetic-geometric transfer, all 0's become 1's and the operation hierarchy moves up by 1 level: Addition to multiplication, multiplication to exponentation, and exponentation to tetration. Therefore, 0^0=1^^1, or 1 tetrated to 1. However, 1 tetrated to 1 is just a power tower of one 1, so if 0^0 is undefined, 1 tetrated to 1 is also indeterminate, and all real and complex numbers are UNDEFINED! Also, 0 factorial has the same discrepancy as we know that factorials of nonnegative numbers is the product of all numbers less than or equal to the number. Since 0 is an annihilator, 0 isn't included in the product. Therefore, 1 is the lowest number that can use the factorial definition. However, we know that (-1)! is infinity, so 0!=(-1!)*0 (due to the factorial rules), but that is infinity*0, which is also undefined. Therefore, 0^0 and 0! live and die together: If 0! is 1, 0^0 is also 1, and if 0^0 is undefined, 0! is also undefined.
In fairness, when considering 0^0, it's kind of important to understand why x^0 is 1 (@1:45) and 0^x is 0.
As for the answer to 0^0, using the limit with real numbers basically definitively denotes 0^0 as 1 (a convergent limit isn't really an "it's close, but I don't know" scenario). The reason 0^0 is considered undefined rather than equal to 1 is, to my understanding, due to indeterminate limits when utilizing complex numbers. So similarly to how 1/x has separate limits for x=0 depending on if approaching from left or right, 0^0 has limits other than 1 is approaching from complex directions.
0^0 comes to a problem of being undefined due to the arithmetic-geometric transfer. In the arithmetic-geometric transfer, all 0's become 1's and the operation hierarchy moves up by 1 level: Addition to multiplication, multiplication to exponentation, and exponentation to tetration. Therefore, 0^0=1^^1, or 1 tetrated to 1. However, 1 tetrated to 1 is just a power tower of one 1, so if 0^0 is undefined, 1 tetrated to 1 is also indeterminate, and all real and complex numbers are UNDEFINED! Also, 0 factorial has the same discrepancy as we know that factorials of nonnegative numbers is the product of all numbers less than or equal to the number. Since 0 is an annihilator, 0 isn't included in the product. Therefore, 1 is the lowest number that can use the factorial definition. However, we know that (-1)! is infinity, so 0!=(-1!)*0 (due to the factorial rules), but that is infinity*0, which is also undefined. Therefore, 0^0 and 0! live and die together: If 0! is 1, 0^0 is also 1, and if 0^0 is undefined, 0! is also undefined.
Eddie Woo 7 years ago had a go at this - I can recommend it to you. The inflection point is very interesting.
0^0 is undefined. According to your situation by using limits, the limit just comes out to be 1. This is according to the graph of y=x^x. Notice that when x=0, that value is undefined and has a point of discontinuity. The limit comes from the use of l'Hopital's rule if you want to know how to get the limit to turn to be 1.
0^0 comes to a problem of being undefined due to the arithmetic-geometric transfer. In the arithmetic-geometric transfer, all 0's become 1's and the operation hierarchy moves up by 1 level: Addition to multiplication, multiplication to exponentation, and exponentation to tetration. Therefore, 0^0=1^^1, or 1 tetrated to 1. However, 1 tetrated to 1 is just a power tower of one 1, so if 0^0 is undefined, 1 tetrated to 1 is also indeterminate, and all real and complex numbers are UNDEFINED! Also, 0 factorial has the same discrepancy as we know that factorials of nonnegative numbers is the product of all numbers less than or equal to the number. Since 0 is an annihilator, 0 isn't included in the product. Therefore, 1 is the lowest number that can use the factorial definition. However, we know that (-1)! is infinity, so 0!=(-1!)*0 (due to the factorial rules), but that is infinity*0, which is also undefined. Therefore, 0^0 and 0! live and die together: If 0! is 1, 0^0 is also 1, and if 0^0 is undefined, 0! is also undefined.
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0^(0) = undefined
0^0 comes to a problem of being undefined due to the arithmetic-geometric transfer. In the arithmetic-geometric transfer, all 0's become 1's and the operation hierarchy moves up by 1 level: Addition to multiplication, multiplication to exponentation, and exponentation to tetration. Therefore, 0^0=1^^1, or 1 tetrated to 1. However, 1 tetrated to 1 is just a power tower of one 1, so if 0^0 is undefined, 1 tetrated to 1 is also indeterminate, and all real and complex numbers are UNDEFINED! Also, 0 factorial has the same discrepancy as we know that factorials of nonnegative numbers is the product of all numbers less than or equal to the number. Since 0 is an annihilator, 0 isn't included in the product. Therefore, 1 is the lowest number that can use the factorial definition. However, we know that (-1)! is infinity, so 0!=(-1!)*0 (due to the factorial rules), but that is infinity*0, which is also undefined. Therefore, 0^0 and 0! live and die together: If 0! is 1, 0^0 is also 1, and if 0^0 is undefined, 0! is also undefined.