c^(infinity)

[Noxide]

I know that 1^(+/- infinity) is indeterminate.
Is c^(+/- infinity) indeterminate for real numbers c?

[jbunniii]

Why do you say that 1^\infty and 1^{-\infty} are indeterminate? We have

\lim_{n \rightarrow \infty} 1^n = 1

and

\lim_{n \rightarrow -\infty} 1^n = 1

If c > 1, then

\lim_{n \rightarrow \infty} c^n = \infty

and

\lim_{n \rightarrow -\infty} c^n = 0

and the opposite result holds for 0 < c < 1

One of the limits doesn't exist if c < -1, and the other one doesn't exist if -1 < c < 0. Neither limit exists if c = -1. (I'll let you work out the details.)

[Mark44]

Think about this in terms of limits. Do you have a feel for what the values of these limits are?
\lim_{n \to \infty} 2^n
\lim_{n \to \infty} 2^{-n} = \lim_{n \to \infty} \frac{1}{2^n}

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.

[Mark44]

Why do you say that 1^\infty and 1^{-\infty} are indeterminate? We have

\lim_{n \rightarrow \infty} 1^n = 1
One of the indeterminate forms is [1^{\infty}]. An example of this is \lim_{x \to 0} (1 + x)^{1/x}.

Here we have the base approaching 1 and the exponent growing large without bound, yet the limit is not 1.


and

\lim_{n \rightarrow -\infty} 1^n = 1

If c > 1, then

\lim_{n \rightarrow \infty} c^n = \infty

and

\lim_{n \rightarrow -\infty} c^n = 0

and the opposite result holds for 0 < c < 1

The limits don't exist if c \leq -1, and they do exist if -1 < c < 0. (I'll let you work out the details.)

[Noxide]

Think about this in terms of limits. Do you have a feel for what the values of these limits are?
\lim_{n \to \infty} 2^n
\lim_{n \to \infty} 2^{-n} = \lim_{n \to \infty} \frac{1}{2^n}

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.

Heh. Gotcha, thanks.