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Why is it indeterminate

  1. Nov 19, 2008 #1
    Can anyone please tell me why 1^(infinity) is regarded to be indeterminate?
    can it be reduced to 0/0? if yes, how?
  2. jcsd
  3. Nov 19, 2008 #2
    Infinity is not a real number. There are no rules for how exponentiation works on something that isn't a number. It cannot be reduced to 0/0, because the expression is essentially meaningless, like asking the IQ of a banana.
  4. Nov 19, 2008 #3
    But still we can reduce forms like (infinity)-(infinity), (infinity)/(infinity), 0*(infinity) etc. to the form 0/0
  5. Nov 19, 2008 #4
    Not under the standard rules of algebra.

    If you see infinity in an expression like that, you're probably seeing it as a shorthand for some kind of limit.

    However, in the case of [tex]1^\infty[/tex], we'd be dealing with something like [tex]\lim_{x \to \infty} 1^x[/tex] which is perfectly reducible to 1, and *not* an indeterminate form.

    And besides that, reducing an expression to 0/0 isn't helpful in the least.
  6. Nov 19, 2008 #5


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    is an indeterminate form because it is undefined (though as mentioned if the 1 is infact 1- we have 0 and if 1 is 1 we get 1 and only when 1 is 1+ doo we have an indeterminate form).
    Suppose we are dealing with limits and not mere expresions in the (affinely) extended real numbers (R*). We can evaluate some limits using the (affinely) extended real numbers .
    lim_{x->infinity} sin(x)/x=0/infinity=0
    works fine
    lim_{x->0} sin(x)/x=0/0
    does not because 0/0 is indeterminate ie undefined
    this does not mean the limit cannot be found
    we might use
    1) Le'hopitals rule
    2) Series expansion
    3) algebraic manipulation

    in particular when using Le'hopitals rule it is handy to exchange one form for another
    1^infinity->exp[ log( 1^infinity)]=exp[infinity*log(1)]->exp[log(1)/(1/infinity)]->exp[0/0]
    lim_{x->infinity} (1+1/n)^n=exp[lim_{x->infinity} log(1+1/n)/(1/n)]
    a 0/0 form
    exp[lim_{x->infinity} log(1+1/n)/(1/n)]=exp[lim_{x->infinity} [log(1+1/n)]'/[(1/n)]']
    =exp[lim_{x->infinity} (-1/n^2)/(1+1/n)/(-1/n^2)]
    =exp[lim_{x->infinity} 1/(1+1/n)]
    =exp[ 1/(1+1/infinity)]
  7. Nov 19, 2008 #6


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    "1/0" is referred to as "undefined" because the equation x= 1/0 is equivalent to 0*x= 1 which is false for all x: 0*x= 0 for all x, not 1. "0/0" is referred to as "indeterminant" because the equation "x= 0/0" is equivalent to 0*x= 0 which is true for all x. 1infinity is "indeterminant" because x= 1infinity is equivalent to ln(x)= infinity*ln(1)= \infinity*0= 0/(1/infinity)= 0/0.

    And that's how you reduce it to 0/0. Of course, what you really reduce is an expression of the form g(x)f(x) where g(x) goes to 1, f(x) goes to infinity. For example, (1/n)n, as n goes to infinity. Do what I suggested before: if y= ((n-1)/n)n= (1-1/n)n as n goes to infinity, then (1- 1/n) goes to 1 while n goes to infinity. Now ln(y)= n ln(1- 1/n)= ln(1- 1/n)/(1/n)= ln(1- m)/m as m goes to 0 the numerator goes to ln(1- 0)= ln(1)= 0 so this is of the form "0/0".
  8. Nov 19, 2008 #7


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    They are indeterminate since they do not have any one particular solution. For example:

    [tex]\lim_{x \to 0} \frac{x^2}{x} = 0 [/tex]


    [tex]\lim_{x \to 0} \frac{x}{x} = 1 [/tex]


    [tex]\lim_{x \to 0} \frac{x}{x^3} = \infty [/tex]

    Note that all of these are of the form 0/0.

    Since the limit of something of the form 0/0 (in this case) does not have any one particular value, it is deemed indeterminate.

    With regard to indeterminate forms, one can reduce a function of the form [itex] 1^\infty [/itex] to a function of the form 0/0.

    Functions of the form 0/0 or [itex] \frac{\infty}{\infty} [/itex] are useful when dealing with limits because one can directly apply L'Hopital's Rule.

    Here's a link discussing it a bit more:


    Hope this helps.

  9. Nov 19, 2008 #8
    So fast and loose with the algebra you are! But I suppose it works for what the OP is probably looking for.

    Using that kind of reasoning, though, [tex]1^\infty = 1[/tex] is just as accurate. For every real x, [tex]1^x = 1[/tex]. It's constant over all the reals, so by extension, it should be constant at both infinities as well.

    To give a more accurate answer, it should be understood that division and exponentiation don't have any one universally accepted definition. You can choose whatever definition you want. Some authors say that 0^0 = 1, and that definitions makes thing like Taylor series a little bit easier, because you can say [tex]exp(x) = \Sigma_{k=0} \frac{x^k}{k!}[/tex] without having to work around the exponent... so we can achieve [tex]exp(0) = 0^0 + 0^1 + 0^2 + \cdots[/tex] without having to make a fuss about the 0^0 term.

    Why isn't 0^0 universally acknowledged to be equal to 1 then? Because then exponentiation wouldn't be continuous! Continuous functions are really nice, and we sort of wish that ^ would be continuous at all points of its domain. But there is no value for 0^0 which can make that so. The result.... we simply say that (0, 0) is not in the domain of ^. It's a continuous function now at the cost that it is no longer defined on all of RxR.

    When you talk about silly things like [tex]1^\infty[/tex], we have similar issues. It's just the opposite situation, actually. We start with a continuous function ([tex]1^x[/tex]) with a restricted domain (all real numbers). By extending the domain to include infinities, we lose continuity.

    But I'm just picky about those sorts of things =-) If [tex]1^\infty = \frac{0}{0}[/tex] seems like it should be true, go for it.
    Last edited: Nov 19, 2008
  10. Nov 21, 2008 #9
    I'm pretty sure what we're dealing here with limits of the form [itex]1^\infty[/itex] is that the 1 is allowed to vary as well, so we're considering the limit [itex]\displaystyle\lim_{(a, b)\to(1, \infty)} a^b[/itex], which doesn't exist: If [itex]t \to \infty[/itex], then [itex]1 + 1/t \to 1[/itex], [itex]t \to \infty[/itex], and [itex]1^t \to 1[/itex] and [itex](1+1/t)^t \to e[/itex], so [itex]1^\infty[/itex] is indeterminate.
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