I know that 1^(+/- infinity) is indeterminate.Is 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.