why infinity^0 is considered as an indeterminate form?
450678
We say that a it is indeterminate because if we have two functions, f and g such that:\n\n   lim (x->oo) f(x) = infinity\n   lim (x->oo) g(x) = 0\n\nthen we know nothing about:\n\n   lim (x->oo) f(x)^g(x)\n\nThat is, this third limit can be any value.\n\nFor example, given a real number A>=1, if we let f(x) = A^x and g(x) = 1/x, then obviously, f(x) converges to infinity, g(x) converges to 0, and f(x)^g(x) converges to A.\n\nIt's important to note that mathematicians aren't trying to be difficult, but often use terms carefully to have a precise meaning.  The phrase 'indeterminate' is different from 'undefined.'\n\nFor the person who says infinity to any power is indeterminate, you are wrong.  The form:\n\n    infinity^R\n\nis determinate if R is a non-zero real number.  If R is negative, then this 'form' is always zero.  If R is positive, this form is always infinity.  This means that if we have two functions f and g so that:\n\n   f(x) -> infinity\n   g(x) -> R\n\nthen f(x)^g(x) -> 0 if R<0, infinity if R>0, and indeterminate if R=0.
997010
x^0 is equal to  x^(1-1) = x^1/x^1 = x/x = 1/1 = 1\n0 is actually undefined. (0^0) because you're dividing by 0.
67585
Formulaically it reduces down to x*z = x*z, so no matter which numbers you plug in for any of the values it will be a valid equation.  \n\n(x*(y+z))-(x*y)=x*z      becomes\nx*y + x*z - x*y = x*z     rewrite that as \nx*y + x*z + -(x*y) = x*z    the positive and negative x*y terms cancel,leaving\n\nx*z = x*z\n\nSince that is a truism, an infinite number of solutions exist.
788532
Is it x*(e-2) or x^(e-2)?\n\nI don't think it matters anyway, because the bottom fluctuates between 3 and 5 while the top increases indefinitely, i.e. approaches "infinity" if you like.  So there's no limit.
95725
Zero of zero is undefined because of the definition of division and some basic field axioms.  It's not zero and it's not infinity, it's UNDEFINED.  I can prove it.\n\nThe multiplicative inverse of n is defined as the number 1/n such that n*(1/n)=1.  It's a property of zero that anything times zero is zero.  Therefore for any 1/n, 0*(1/n)=0, and since zero cannot equal one, zero has no multiplicative inverse.\n\nDivision by n is defined as multiplication by the multiplicative inverse of n.  Since zero has no multiplicative inverse, you cannot divide by zero.\n\nAll you people who said that anything divide by zero is infinity are just flat out wrong.  If n is not equal to zero, the limit as x approaches zero for the positive side of n/x is infinity, but that does not mean that n/0 is infinity.  Also, note that if you come at it from the negative side the limit is negative infinity.  n/0 is undefined because zero has no mutiplicative inverse so the expression 1/0 is mathematically meaningless  Also, the limit of 0/x as x approaches zero would be zero, not infinity.\n\nWhoever it was who did the calculator thing, calculators are a far cry from a mathemtical proof.   There are plenty of arithmetic problems that make perfect sense but can't be done on a calculator.  In fact, there are infinetly many of them.  And there are some properties of multiplication and addition that can be proven mathematically but are not true on a calculator.  But you and the first poster get major props for being the only people to answer this question and demonstrate anything resembing an basic understand of elementary arithmetic.\n\n______\nH2++ cannot exist because it would have to have a covalent bond but would have no electrons.  Two hydrogen nuclei can't form a diatomic molecule.  O-- is not going to exist in water because the hyrogen atom would have to dissociate from the water molecule one at a time, and OH- is more stable than O-- and H+.  It could in theory exist as 2H+ and O--, but OH- is a much happier molecule than O--.\n\nNothing can go faster than the speed of light because as an object approaches the speed of light it's mass becomes infinite, and so the force required to cause it to accelerator more becomes infinite.  You can't apply infinite force to something.  Normally I'd do out the math to show that, but it's 3:30 in the morning so I doubt I can make it coherent.\n\nTensor calculus is too complicated to explain at 3:30 in the morning.  You can read about it here: http://en.wikipedia.org/wiki/Tensor_calculus
687976
The ideas of zero and infinity are really difficult, mainly because we have hardly, if ever, seen them in our lives. When have you really seen nothing? Never. When have you seen everything? Never. \n\nAnd a quantity of infinity does not mean it goes on forever, that is a mis-conception. Quantities dont go, rates do.\nInfinity is the opposite of nothing. Its everything. Not like everything in front of you, that is a limited number. But EVERYTHING. The maximum of existance.
892605
(constant)/(2^n) when n->infinity => the limit is 0(zero)\n\nPS\nNot to be a a stickler for rules but is this really laptop related ? :))
532480
Undefined, again the common mistake is to treat infinity as a number.  Infinity does not play with the same rules as numbers.
487867
Here is the mathematical proof:\n\nIt involves some precalculus, so I will explain what I am doing just in case you haven't gotten to precalc yet.\n\n[1]  Definition: Limit\n     The limit of a function f(x) is the value of y @ x.\n     i.e. y = 2x^2, @ x = 1, y = 2, \n     Therefore, \n     limit 2x^2 = 2\n     x->1\n\n[2]  Why division by Zero = infinity:\n\n     limit 1/x = infinity\n     x -> +0\n\n     Limit 1/x = infinity\n     x -> -0\n\nThe graph of y = 1/x explains this principle. \n\n[2.1] Definition: +0  & -0\n\n+0 = As x approaches zero from the right.\n-0 = As x approaches zero from the left.\n\n[2.2] As x get closer and closer to zero, the quotient of 1/x gets larger and larger.\n\ni.e. 1/0.0000006 = 6000000\ni.e  1/1x10^-9999= 1x10^9999\n\n[2.3] When x becomes zero, the quotient of 1/x is infinity. Since infinity is underfined, 1/0 is also undefined!
665836
This question already discussed here so many times. For your information the answer is: "Indeterminate". \n\nTo say that "0/0" is an indeterminate form does not just mean that "0/0" by itself can represent any number, or may represent no number. Those points are true, in a certain sense, but of limited practical significance when stated in those terms. Unlike "0/0", "1/0" is not an indeterminate form.\n\nIt means also that the ratio of two functions f and g that approach zero can approach any member of a range of well-defined values, depending on which functions f and g are. Whether such a value exists, and what it might be, depends on how the functions approach zero.\n\nIn more formal language, the fact that the functions f(x) and g(x) both approach 0 as x approaches some limit c, is not enough information to evaluate the limit\n\nlim f(x)/g(x)\nx->c\n \nThat limit could be any number or plus or minus infinity, or might not exist, depending on what the functions f and g are.\n\nBy contrast "1/0" is not an indeterminate form because there is no range of different values that f/g could approach if f approaches 1 and g approaches 0. (The absolute value of the ratio will always approach positive infinite and so is not considered indeterminate.)
306774
Anything divided by 0 is actually undefined, not infinity or 0. When you think about how that works, it makes sense:\n\nDividing a numerator by smaller numbers makes your number larger. Divide 4 by 2, you get 2. Divide 4 by 1, you get 4. Divide 4 by 0.5, you get 8. And so on until you are dividing by a number so small it's pretty close to 0. The overall number gets larger and goes off into infinity.\n\nBut, if you divide a negative number in the same manner, your number gets smaller. -4 by 2 is -2, -4 by 1 is -4, -4 by 0.5 is -8, etc. As you approach 0, your numbers go off into negative infinity.\n\nYou'll have the same weird effect if you divide by negative numbers - a positive divided by an increasingly smaller negative will yield negative infinity, and a negative divided by an increasingly smaller negative will yield positive infinity.\n\nYou can't have it both ways - dividing by zero yields both a positive and negative infinity! So someone a long time ago decided that since a number can't really have two values, it will be undefined.
241288
What goes wrong is the loss of information.  All _n_ divided by zero goes into the same "infinite quality".\n\nIt is possible to admit different infinite numbers; this is Robinson's approach with "nonstandard analysis", and then there are a myriad of choices as what to admit and what to disallow.  If you take that approach, it is still easier to disallow n/0 as undefined, and permit other "zeroes" where n/d is defined, where d is one of those classes of zeroes.\n\nIt is exactly equivalent, however, to the following:\nlim(d->0) n/d.  In this case the limit doesn't exist, but there are other "zero approaching denomenators" where the limit does exist.  One very important one:\n\nlim(d->0) sin(d)/d = 1
208997
The only way to get the expression infinity/-infinity is through a limiting process in the real numbers, so I'll assume you're talking about limits. In that case the form above is indeterminate, meaning that it depends on the functional forms in the limit. \nSuppose you have f(x) and g(x) such that lim [f(x)/g(x)] = infty/-infty. Consider the equivalent expression lim [f(x)/g(x)] = lim [(1/g(x))/(1/f(x))] which approaches the form 0/0. This allows the use of L'Hospital's rule to evaluate the limit, which we know from experience can be anything.
8320
0 is non-negative, since all negative are < 0\n0 is not pozitive, since all pozitive are > 0\n0 is Natural\n0 is Integer\n0 is Fractional as 0/x\n0 is Rational\n0 is Real\n0 is Complex\n0 is Finite\n\n0 + x = x | for every x of A
490253
infinity/0 is infinity.  If you have any function f(x) and g(x) where f(x) goes to infinity as x -> k and g(x) goes to zero as x -> k, then 1/g(x) goes to infinity as x -> k, so\n\nf(x)*(1/g(x)) -> infinity as x -> k.\n\ninfinity*zero depends. If you look at 1/x and x^2, then as x -> 0, 1/x -> infinity and x^2 -> 0, but\n\n(1/x)*(x^2) = x, so its limit is 0 as x -> 0.\n\nIf you look at 1/x^2 and x, then as x -> 0, 1/x^2 -> infinity and x -> 0, but\n\n(1/x^2)*(x) = 1/x, so its limit is infinity as x -> 0.
1187309
I think I did it.  If y=1/log(x), then a few manipulations gave me\n\nx=10^(1/y).\n\nI'm assuming by log you mean log to the base 10.  I suppose mathematicians will start fussing about what happens at y=0, but a physicist like me doesn't worry that much about infinities ...
407128
1^infinity = infinity\n1/1^infinity = 0\nthere 1/0 = 1 ^infinity\n THATS IT , I AM TURNING TO BE A GENIUS