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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.
Associated Topics:
High School Calculus
High School Definitions
High School Functions
High School Polynomials

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