0
$\begingroup$

Is it really true that $$\lim_{n\to\infty} 1^n = 1$$

?

We were taught throughout our entire math courses that $[1^{\infty}]$ is an undetermined form....Am I missing something here?

$\endgroup$

marked as duplicate by GNUSupporter 8964民主女神 地下教會, jvdhooft, Namaste calculus Mar 30 '18 at 14:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ yes, $\lim_{n\to\infty}1=1$. $\endgroup$ – Lord Shark the Unknown Mar 30 '18 at 11:42
  • 4
    $\begingroup$ Actually, $\lim_{x\to\infty}1^n=1^n$. $\endgroup$ – José Carlos Santos Mar 30 '18 at 11:42
5
$\begingroup$

Is it really true that $$\lim_{\color{red}{x}\to\infty} 1^n = 1$$

You probably mean: $$\lim_{\color{blue}{n}\to\infty} 1^n = 1$$ and yes, this is true because $1^n = 1$ for all $n$.

The expression $"1^{+\infty}"$ is indeterminate and the limit above doesn't contradict that.

Perhaps you know the following well-known limit too: $$\lim_{n\to\infty}\left(\color{blue}{1+\frac{1}{n}}\right)^\color{red}{n} = e \ne1$$where you also have $\color{blue}{1+\frac{1}{n}}\to 1$ and $\color{red}{n}\to+\infty$.

Combining both limits shows that you can have sequences $c_n = \left(a_n\right)^{b_n}$ where $a_n \to 1$ and $b_n\to+\infty$ but with different limits for $c_n$; which is why we call $"1^{+\infty}"$ indeterminate.

$\endgroup$
3
$\begingroup$

You're clearly misunderstanding what an "indeterminate form" is. The fact that $1^\infty$ is an indeterminate form means that knowing $f(n)\to1$ and $g(n)\to \infty$ does not allow us to say anything in general about $$\lim_n f(n)^{g(n)}$$

However, this does not mean that limits of this kind cannot exist. Examples: $$\lim\limits_{n\to\infty} \left(1+\frac1n\right)^n = e $$ and your case $$\lim\limits_{n\to\infty}1^n = 1. $$

$\endgroup$
1
$\begingroup$

$$\lim_{n\rightarrow \infty}{1^n}=\lim_{n\rightarrow \infty}{1}=1$$ Since $1^n=1 \ \forall n$, thus the limit is now independent of n and thus stays the same.

$\endgroup$
1
$\begingroup$

You have to distinguish from purely a constant base $1$ to a power approaching infinity and a base that is a function approaching $1$ while its power approaches infinity (and dependent on the same limiting variable).

A constant can never change. A function can. This impacts your end result. Take the famous limit:

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

Realize the base depends on $n$ to approach $1$ while the power also depends on $n$. Thus, the 'different rates of approachment' lead to unexpected results that make it indeterminate.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.