prove that -  1/0 = infinity?
407128
1^infinity = infinity\n1/1^infinity = 0\nthere 1/0 = 1 ^infinity\n THATS IT , I AM TURNING TO BE A GENIUS
721127
it's the same as infinity plus infinity infinity times, of course.  ;)\n\nnot to be confused with infinity minus infinity infinity times, which would be the opposite.\n\nalso not to be confused with infinity divided by infinity infinity times, which would be a positive number as close to zero as possible w/o reaching zero.
596523
If you are approaching zero from the left, then -1 divided by 0 is positive infinity.\nIf you are approaching zero from the right, then -1 divided by 0 is negative infinity.\n-1 divided by 0 is undefined because you get different results when coming in from the right side of 0 and the left side of 0.\nI learned this in Calc 1 (possibly Calc 2).\nThere are different levels (degrees) of infinity.
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!
566883
First, since 1/x is not defined for x = 0, before you can show  anything about your functions, you have to define values for g(0) and f(0). I'm going to assume that either the original statement of the problem had that or that you are supposed to find the value that makes the problem solvable.\n\nTo be true you have to define the problems with f(x) = 0 and g(x) = 0\n\n\nFor the continutiy of f(x) given epsilon > 0 find delta > 0 such that |f(x) - f(0)| < epsilon for all |x-0| < delta.\n\n|f(x) - f(0)| = |xsin(1/x) - 0| < |x| = |x - 0|\n\nGiven epsilon, pick delta = epsilon.\nIf |x-0| < delta, then |f(x) - f(0)| < epsilon.\n\n\nAs for the differentiability of g(x),\n\nf(0+h) - f(0)) = (0+h)^2sin(1/(0+h)) - 0\n\nlim(x->0)(f(0+h) - f(0))/h = \nlim(x->0)((0+h)^2sin(1/(0+h)) - 0)/h = \nlim(x->0) h^2(sin(1/h)/h = lim hsin(1/h) = 0\n\nSo the limit exists and is 0.
98782
Just substituting x for 0 yields -9/-3 = 3.  This is not the indeterminate form 0/0 hence the limit is obtained by substitution, ie the answer is 3.
494525
Think of it in terms of logarithms, since it is the inverse of exponents.\n\nStatement to prove:  1^infinity = 1\n\nRewritten in logarthmic form:  log[1](1) = infinity\n(Statement above is read: log base 1 of 1 equals infinity)\n\nWell that can't make sense, right?  It is a law of logs and of exponents that log[b](1) = 0 for any positive base b.  This is the same as saying b^0 = 1, for any b.  Also, log[b](b) = 1, for any positive b.  This makes since in b^1 = b.  Both of these laws would conflict this reasoning.  In fact, when b = 1, they conflict with one another.  Therefore, lets look at it with a variable instead.\n\nProve:  1^infinity = X\n\nRewrite in logartithmic form: log[1](X) = infinity\n\nNow, what value of X will give you infinity?\nCan you determine that?\n\nI would say it is indeterminate :)
565951
the answer is zero b'coz at the end, X-X is equal to zero.
1137788
Nope, 0/0 could be 1 but it could also be anything else.   Division is the opposite of multiplication and 0 times anything at all, even infinity, is still zero.
344009
No, because it doesn't.\nThere's a number of "proofs" out there that rely on dividing by 0, even when it's not obvious that's what's going on.\nDivide by zero, and you can prove anything!
497395
1.\nBecause e to the power minus infinity is 0.
1154919
<1>...u know lim xsin(1/x)=lim (sin(1/x))/(1/x),and when x goes to infinity ,(1/x) goes to <0>....and we know that (sina)/a when <a> goes to <0> is 1...and here <a> is <1/x>...so it's limit is <1>
646401
it's the inverse x^-1=1/x
932401
the answer is 0\n\nu know, lim 1/(n+1) = 0\n              n -> ∞ \n\nand, lim (n/n+1)^n = 1\n        n -> ∞ \n\nso: lim (1/n+1)(n/n+1)^n = lim (1/n+1) * lim(n/n+1)^n = 0 * 1 = 0\n      n -> ∞                           n -> ∞          n -> ∞
574280
a number to the zero exponent is always 1 as a answer
870318
infinity could be ANY number\n\nif infinity(1) is 32,767 and infinity(2) is 868, the division does not equal one.  I just chose random numbers for that example.
858967
Division by zero is an "illegal" operation. Some claim it's infinity because 142/x goes towards infinity as x goes towards zero, but that's not the same as 142/0 = infinity.\n\nLuckily, you never need to divide by zero for any practical purposes, so it's not a problem.