The discussion centers on the indeterminacy of expressions involving exponentiation, specifically questioning whether \( c^{(+/- \infty)} \) is indeterminate for real numbers \( c \). Participants explore the behavior of limits as the exponent approaches positive or negative infinity, comparing it to the known case of \( 1^{(+/- \infty)} \).
Participants express differing views on the indeterminacy of \( c^{(+/- \infty)} \) based on the value of \( c \). There is no consensus on whether \( 1^{(+/- \infty)} \) should be classified as indeterminate, as some participants provide counterexamples and alternative interpretations.
Participants acknowledge that the behavior of limits can depend on the specific values of \( c \) and the conditions under which they are evaluated, but do not resolve these complexities.
One of the indeterminate forms is [itex][1^{\infty}][/itex]. An example of this is [tex]\lim_{x \to 0} (1 + x)^{1/x}[/tex].jbunniii said:Why do you say that [itex]1^\infty[/itex] and [itex]1^{-\infty}[/itex] are indeterminate? We have
[tex]\lim_{n \rightarrow \infty} 1^n = 1[/tex]
jbunniii said:and
[tex]\lim_{n \rightarrow -\infty} 1^n = 1[/tex]
If [itex]c > 1[/itex], then
[tex]\lim_{n \rightarrow \infty} c^n = \infty[/tex]
and
[tex]\lim_{n \rightarrow -\infty} c^n = 0[/tex]
and the opposite result holds for [itex]0 < c < 1[/itex]
The limits don't exist if [itex]c \leq -1[/itex], and they do exist if [itex]-1 < c < 0[/itex]. (I'll let you work out the details.)
Mark44 said:Think about this in terms of limits. Do you have a feel for what the values of these limits are?
[tex]\lim_{n \to \infty} 2^n[/tex]
[tex]\lim_{n \to \infty} 2^{-n} = \lim_{n \to \infty} \frac{1}{2^n}[/tex]
To investigate this further for arbitrary bases, you probably want to limit the base to the positive reals. One case would be for a > 1. Another would be for 0 < a < 1.