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

this is probably a very simple question for analysts, but i don't understand why the limit of the function $ \lim_{n \to \infty}1^n$ does not exist.

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    $\begingroup$ It is defined. This isn't what people mean when they say that $1^{\infty}$ is an indeterminate form. $\endgroup$ – Qiaochu Yuan Aug 5 '11 at 16:03
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    $\begingroup$ Just to be clear: The term $1^n = 1$ for all $n$. So, $\displaystyle\lim_{n \to \infty} 1^n = 1$. However, if $\displaystyle\lim_{n \to \infty} f(n) = 1$ and $\displaystyle\lim_{n \to \infty} g(n) = \infty$, then $\displaystyle\lim_{n \to \infty} f(n) ^{g(n)}$ is indeterminant. This last fact is discussed at length in the link above. $\endgroup$ – JavaMan Aug 5 '11 at 16:09

Let $f_1(n)=1^n.$ Then $\displaystyle\lim_{n\to\infty}f_1(n)=1$.

So there is no problem in evaluating $\displaystyle\lim_{n\to\infty} 1^n$.

Let $f_2(n)=\left(1+\frac{1}{n}\right)^n.$

Then $\displaystyle\lim_{n\to\infty}f_2(n)=e$.

Thus the difficulties come when we raise to the $n$-th power something that is approaching $1$.

Here are a few more examples.

Let $f_3(n)=\left(1+\frac{1}{\sqrt{n}}\right)^n$. Then $\displaystyle\lim_{n\to\infty} f_3(n)$ does not exist, or if you prefer, is $+\infty$.

Let $f_4(n)=\left(1-\frac{1}{\sqrt{n}}\right)^n$. Then $\displaystyle\lim_{n\to\infty} f_4(n)=0$.

Here is a slightly more complicated example. Let $f_5(n)=\left(1+\frac{\sin^2(\pi n/2)}{n}\right)^n$.

Then, for large $n$, $f_5(n)$ is bouncing around. Sometimes it is $1$, sometimes it is close to $e$.

We can produce examples that exhibit much weirder behaviour!


It is defined. It's equal to $1$.


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