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

1. Jul 7, 2010

### Noxide

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

2. Jul 7, 2010

### jbunniii

Re: c^(infinity)

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.)

3. Jul 7, 2010

### Staff: Mentor

Re: c^(infinity)

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.

4. Jul 7, 2010

### Staff: Mentor

Re: c^(infinity)

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.

5. Jul 7, 2010

### Noxide

Re: c^(infinity)

Heh. Gotcha, thanks.