Undefined and Indeterminable ... at the Same Time?
Date: 09/05/2010 at 20:16:37 From: Michael Subject: "Indeterminate Form" and "Undefined" Mutually exclusive? Are the terms "indeterminate form" and "undefined/no solution" mutually exclusive? In other words, can an expression be considered both an indeterminate form AND undefined? Of course it could be neither -- but is it only ever one or the other? I know it is possible for an expression to be ONLY undefined, as is 1/0. But can something like 0/0 or 0^0 be considered both undefined and indeterminate? Or are they just indeterminate?
Date: 09/05/2010 at 20:34:29 From: Doctor Vogler Subject: Re: "Indeterminate Form" and "Undefined" Mutually exclusive? Hi Michael, Thanks for writing to Dr. Math. The phrase "indeterminate form" is used in the context of limits, whereas "undefined" refers to evaluating functions, and "no solution" refers to solving equations or similar problems. Let's look at some examples from each of these different contexts. Consider the function f(x) = (x^2 - 4)/(x - 2) It is undefined at x = 2, even though it has a limit as x -> 2. Indeed, the limit has the indeterminate form 0/0 and therefore one way to evaluate the limit is using L'Hopital's Rule. (Another way is dividing out the common factor.) Also, the limit ... lim (1 + 1/x)^x x->infinity ... has the indeterminate form 1^infinity, even though infinity is not a number, so it makes no sense to "evaluate" this function at infinity (except as taking the limit). Therefore, saying that it is "undefined" at infinity is rather meaningless. But if you are taking a limit at a finite point (not a limit at infinity), then you can relate evaluating the limit as x -> c to evaluating the function at c. In this context, you can compare "undefined" to "indeterminate." I should warn you that the limit as x -> c does not depend on the value of the function at c. However, indeterminate forms apply when the function in question is some operation (like the quotient) of two *continuous* functions at c. So, for example, we say that 0/0 is an indeterminate form because if ... f(x) = g(x)/h(x) ... for continuous functions g and h (at the point c) and g(c) = h(c) = 0, this is not enough information to determine the limit of f(x) as x -> c. But it is enough information if, for example, h(c) = g(c) = 1. Normally, it is true that if your limit has an indeterminate form, then evaluating the function at that point will yield an undefined result, in the sense of the continuous functions referred to above. However, I can think of one exception to this rule: the last example you named! For several reasons, the expression 0^0 is generally understood to evaluate to 1, even though (in limits) this is an indeterminate form, because as x -> 0, 0^(x^2) -> 0 and x^0 -> 1. Of course, a function with a "removable discontinuity" is undefined at the point in question even if the limit does not have an indeterminate form. So in that sense, the two concepts are not really related. Finally, I would say that this limit does not exist: lim cos(1/x) x->0 By comparison, this equation has no solution: x^2 + 1 = (x - 1)(x + 1) Does that clear things up? If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/
Date: 09/05/2010 at 20:38:51 From: Michael Subject: "Indeterminate Form" and "Undefined" Mutually exclusive? Thanks, I understand it now.
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