Regarding limits and $1^\infty$ [duplicate]

Possible Duplicate:
Why is $1^{\infty}$ considered to be an indeterminate form

I have some questions about limits and the undefinability of $1^\infty$.

For example, is $\lim_{x\to\infty}1^x$ indefinite? Why is it not $1$? Or do mathematicians, when saying that $1^\infty$ is indefinite, actually refer to cases such as $lim_{x\to\infty} \left(1 + \frac{a}{x}\right)^x$ where even though at a first glace the result is $1$, this is actually a special case and it is equal to $e^a$?

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marked as duplicate by Chandru, Qiaochu YuanMay 21 '11 at 18:29

Please see the reference. –  user9413 May 21 '11 at 18:26
@Chandru I'm going through their answers right now. :) –  Paul Manta May 21 '11 at 18:29
Sure, go through. –  user9413 May 21 '11 at 18:30

When people say that $1^{\infty}$ is an indeterminate form, what they mean is that if $f(x)$ is a function such that $\lim_{x \to r} f(x) = 1$ and $g(x)$ is a function such that $\lim_{x \to r} g(x) = \infty$, the value of $\lim_{x \to r} f(x)^{g(x)}$ is not uniquely determined in general. It is determined in the special case that $f(x) = 1$, in which case the limit is obviously just $1$.

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This is a duplicate question. –  user9413 May 21 '11 at 18:27

By itself, $1^{\infty}$ doesn't mean anything, except the implication that you had a quantity of the form $a^b$, and $\lim_{x\rightarrow\infty}a=1$ and $\lim_{x\rightarrow\infty}b=\infty$. Depending on how quickly and from what direction $a$ approaches 1, and how quickly $b$ approaches $\infty$, $a^b$ could approach any non-negative number, infinity, or nothing in particular at all.

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Look up indeterminate forms. It depends on how you interpret $1^{\infty}$

The indeterminate form $1^{\infty} = \lim_{n \rightarrow \infty} f(n)^{g(n)}$ where $\lim_{n \rightarrow \infty} f(n) = 1$ and $\lim_{n \rightarrow \infty} g(n) = \infty$

For instance, all of the following limits are of the form $1^{\infty}$, yet they all evaluate to different numbers

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

$$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{n} \right)^n = e$$

$$\lim_{n \rightarrow \infty} \left(1 + \frac{1}{\log(n)} \right)^n = \infty$$

$$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{\log(n)} \right)^n = 0$$

It is a question of which one tends faster whether $f(n) \rightarrow 1$ or $g(n) \rightarrow \infty$.

When you get a finite number which is non-zero and not one as the limit, there is in some sense a balance between the rate at which $f(n) \rightarrow 1$ and $g(n) \rightarrow \infty$.

When you get infinity or zero as the limit, $g(n) \rightarrow \infty$ faster than $f(n) \rightarrow 1$.

When you get $1$ as the limit, $f(n) \rightarrow 1$ faster than $g(n) \rightarrow \infty$.

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