Why Are 1^infinity, infinity^0, and 0^0 Indeterminate Forms?Date: 05/08/98 at 11:44:56 From: Tim Hanna Subject: Indeterminates I am a senior in Advanced Placement Calculus, and my class is having a hard time understanding some indeterminate forms. We know that these are indeterminate forms: 0/0 infinity/infinity infinity - infinity, But why are these indeterminate forms? 1^infinity infinity^0 0^0 We feel that 1^infinity = 1, infinity^0 = 1, and 0^0 = 1. Part of these conclusions come from the fact that 0^infinity = 0 and 0^(- infinity) = infinity. Could you please explain these determinate and indeterminate forms? Date: 05/08/98 at 15:31:02 From: Doctor Rob Subject: Re: Indeterminates These forms are called indeterminate because if you replace 1, 0, and infinity by functions the limits of which are 1, 0, and infinity as x -> 0, then the limit of the compound function does not exist, in the sense that the limit depends on which functions you choose. For an example of this discrepancy for 1^infinity, on the one hand, take f(x) = 1 and g(x) = 1/x. Then: lim f(x)^g(x) = lim 1^(1/x) = lim 1 = 1 On the other hand, if we take f(x) = 1 + x and g(x) = 1/x, then: lim f(x)^g(x) = lim (1 + x)^(1/x) = e = 2.718281828459... > 1 For an example of 0^0, on the one hand, take f(x) = 0, g(x) = x. Then: lim f(x)^g(x) = lim 0^x = lim 0 = 0 On the other hand, if we take f(x) = x, and g(x) = 0, then: lim f(x)^g(x) = lim x^0 = lim 1 = 1 > 0 All of the seven indeterminate forms are the same: 0/0 [(k*x)/x -> k, for any real k] 0*infinity [(k*x)*(1/x) -> k, for any real k] infinity/infinity [(k/x)/(1/x) -> k for any real k] infinity - infinity [(k+1/x)-(1/x) -> k for any real k] 1^infinity [(1+x)^(ln[k]/x) -> k for any positive real k] infinity^0 [(1/x)^(ln[k]*x/[1-x]) -> k for any positive real k] 0^0 [x^(ln[k]/ln[x]) -> k for any positive real k] It is a useful exercise to prove all these statements. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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