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First of all, I'm sorry if this question has been already asked and answered, as far as I searched, I couldn't find such a question on this site. So, I've been thinking about the limit of the sequence $\left(\root{2n+1}\of{-1}\right)_{n\geq 0}$. Since the order of the root is odd for every $n$, this sequence, is obviously a constant sequence with the general term $a_n = -1$. So, from this follows that $$ \lim_{n\to\infty} \root{2n+1}\of{-1} = -1 $$. We can even do an $\epsilon-N$ proof to show this (and it's realy easy actually): $$ \forall \epsilon > 0 \hspace{0.5cm} \exists N \geq 0 \hspace{0.3cm} \text{s.t.} \hspace{0.3cm} \left|\root{2n+1}\of{-1}+1\right|<\epsilon \hspace{0.5cm} \forall n \geq N \\ \left|\root{2n+1}\of{-1} + 1\right| = \left|-1 + 1\right| = 0 < \epsilon \hspace{0.5cm} \forall n \geq 0 \\ N = 0 \ _\blacksquare $$.

However, if we use tehniques ussualy used for solving limits, we end up with a different result: $$ \begin{align*} \lim_{n\to\infty} \root{2n+1}\of{-1} &= \lim_{n\to\infty} (-1)^{1\over 2n+1} \\ &= \left(\lim_{n\to\infty}-1\right)^{\lim_{n\to\infty}{1\over 2n+1}} \\ &= (-1)^0 \\ &= 1 \end{align*} $$.

What is wrong here? Why do the two methods give different results?

Edit: To make everything clear, I'm assuming the real root as defined by $\root n \of {} : \mathbb{R} \to \mathbb{R}$ for odd $n$ and treating this as a real-analysis problem. Also, it's pretty explicit from my question that I'm working with a sequence and not with a function. The limit only goes through natural values of $n$

Edit 2: I've figured it out. Thank you all for your answers esspecially to @Jack who pointed out the theorem I've been using $\lim_{n\to\infty}(a_n^{b_n}) = (\lim_{n\to\infty} a_n)^{(\lim_{n\to\infty} b_n)}$ is not true in general. I've consulted my textbook again and saw that I've missed the part where they said $a_n > 0, \forall n \in \mathbb{N}$. Of course, we can think of this problem also from the viewpoint of functions and the fact that the function $(-1)^x$ is not continuous is another gap in using something like the above theorem. Thank you all again for being so kind and giving me so many answers.

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    $\begingroup$ The problem is that you need to pick a definition of what $\sqrt[n]{z}$ means for a complex number $z$. This function will have discontinuities. $\endgroup$ – Carlos Esparza Sep 26 at 14:24
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    $\begingroup$ Your problem is not well-defined. With positive real numbers, $a^{1/n}$ means the positive $n^{\rm th}$ root. But with a non-positive number, there are $n$ choices for an $n^{\rm th}$ root of $a$, and no privileged. one. $\endgroup$ – Martin Argerami Sep 26 at 14:25
  • $\begingroup$ Why is $\sqrt[2n+1]{-1}=-1$ for every $n\ge 0$? $\endgroup$ – Bman72 Sep 26 at 14:27
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    $\begingroup$ @StefanOctavian: what book are you using? $\endgroup$ – Jack Sep 26 at 14:37
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    $\begingroup$ Yes, that is the book. I misread theorem 13 at page 133 $\endgroup$ – Stefan Octavian Sep 27 at 9:21
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Your expression $\sqrt[2n+1]{-1}$ (for any nonnegative integers $n$) is defined to be, as you stated in the post, the unique real number $y$ such that $y^{n+1}=-1$. Since by your definition, $\sqrt[2n+1]{-1}=-1$, there is no doubt that $$ \lim_{n\to\infty}\sqrt[2n+1]{-1}=\lim_{n\to\infty}(-1)=-1. $$

There is no problem for the limit itself.

What goes wrong here is in your second "method":

if we use techniques usually used for solving limits, we end up with a different result: $$ \begin{align*} \lim_{n\to\infty} \root{2n+1}\of{-1} &= \lim_{n\to\infty} (-1)^{1\over 2n+1} \\ &= \left(\lim_{n\to\infty}-1\right)^{\lim_{n\to\infty}{1\over 2n+1}} \\ &= (-1)^0 \\ &= 1 \end{align*} $$.

The following step is problematic: $$ \lim_{n\to\infty} (-1)^{1\over 2n+1} = \left(\lim_{n\to\infty}-1\right)^{\lim_{n\to\infty}{1\over 2n+1}} $$

What you use here is $$ \lim_{n\to\infty}{a_n}^{b_n}=(\lim_{n\to\infty}a_n)^{(\lim_{n\to\infty}b_n)} \tag{1} $$ where $a_n=-1$ is the constant sequence and $b_n=\frac{1}{2n+1}$. But (1) is NOT true in general.


[Added] In real analysis, one rarely writes expression like $a^b$ for $a\leq 0$ and arbitrary real number $b$, unless one specifically defines such expression for some particular $a$ and $b$. For instance, you define $(-1)^{1/n}$ for only $n$ being an odd positive integer and let $(-1)^{1/n}$ be the unique number $y$ such that $y^{n}=-1$. In such situation, $(-1)^{1/n}$ is nothing but the real number $-1$.

One definition for the expression $a^b$ with $a>0$ and $b\in\mathbb{R}$ is $e^{b\ln a}$. And one has the following statement

Suppose $\{a_n\}$ is a positive sequence of real numbers such that $\lim_{n\to \infty}a_n=a$. Assume in addition that $\{b_n\}$ is a real sequence with $\lim_{n\to\infty}b_n=b$. Then $$ \lim_{n\to \infty}a_n^{b_n}=\lim_{n\to \infty} e^{b_n\ln a_n}=\lim_{n\to\infty}e^{b\ln a}=a^b. $$

If one does want to consider the expression $a^b$ for negative real number $a$, then one would

  • either stick to the definition for the some specific $a$ one has,

  • or unavoidably talk about the complex logarithm. See also this Wikipedia article.

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  • $\begingroup$ thank you so much for your answer. Can you tell me some conditions for which (1) is true or direct me to some resources? $\endgroup$ – Stefan Octavian Sep 26 at 14:58
  • $\begingroup$ @StefanOctavian: you are welcome. I have added a few words into the answer. $\endgroup$ – Jack Sep 26 at 15:42
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As you know $$\root{2n+1}\of{-1}$$ is not just one number but $2n+1$ different numbers.

Thus the following limit is not even well-posed.

$$\lim_{n\to\infty} \root{2n+1}\of{-1} $$

Of course $-1$ is always included in the set of $2n+1^{st}$ roots of $-1$ so if you choose that root for every $2n+1$, then you may say that $$\lim_{n\to\infty} \root{2n+1}\of{-1}=-1 $$

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  • $\begingroup$ I'm assuming principal root $\endgroup$ – Stefan Octavian Sep 26 at 14:30
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    $\begingroup$ @stefanoctavian But the principal root does tend to one... $\endgroup$ – Peter Foreman Sep 26 at 14:35
  • $\begingroup$ I'm sorry. I meant real root not principal root $\endgroup$ – Stefan Octavian Sep 26 at 14:42
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There are two ways of looking at this.

The first is that you require the root to be real. In that case, both methods give -1. Because there is only one real root for any $2n+1$ and it is -1. So the 2nd line of your final 4 line derivation is wrong: the rhs is simply $\lim_{n\to\infty}-1=-1$.

The other way of looking at it is that we really need to work with complex numbers to figure out what is going on. In that case there are $2n+1$ roots, so you have to decide which one you are picking when you take the limit. They are all on the unit circle - are you familiar with the Argand diagram? If you pick the one with the smallest "argument" (ie angle to the positive real axis) each time, then you get +1 as the limit. If you pick the one with the largest angle each time, then you get -1 as the limit.

Incidentally, I do not understand your $\epsilon-\delta$ proof.

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I think the problem lies in this step: $$\lim_{n\to\infty} (-1)^{1\over 2n+1}= \left(\lim_{n\to\infty}-1\right)^{\lim_{n\to\infty}{1\over 2n+1}}$$

You seem to be using, that the function $a, b \mapsto a^b$ is continuous. But in real Analysis this function is only defined when $b = \frac1{m}$ for some integer $m$. Therefore you cannot interchange the limits.

(Of course you can define $a, b \mapsto a^b$ as a continuous function but then you need to use complex analysis and will run into the discontinuities I mentioned in my comment.)

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  • $\begingroup$ Please read my question carefully. I'm explicitly talking about a sequence, not a function. The limit is only for $n \in \mathbb{N}$ $\endgroup$ – Stefan Octavian Sep 26 at 14:49
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    $\begingroup$ @StefanOctavian Yes, that's the thing: Interchanging the limits like $\lim_{n \to \infty} (-1)^{\frac1{2n + 1}} = (-1)^{\lim_{n \to \infty}\frac1{2n+1}}$ is only valid if $x \mapsto (-1)^{x}$ is a continuous function. $\endgroup$ – Carlos Esparza Sep 26 at 14:52
  • $\begingroup$ Sequences are functions @StefanOctavian $\endgroup$ – Bman72 Sep 26 at 15:05
  • $\begingroup$ @0x539 Sequences remain functions. You can use the rule that you used, only if the limit of the base sequence and the one of the power sequence exists and if the function $(-1)^x$ is continuous. This doesn't imply that sequences aren't functions. $\endgroup$ – Bman72 Sep 26 at 15:13
  • $\begingroup$ @Bman72, sequences are a special type of functions: they are defined on some subset of $\mathbb{Z}$ usually $\mathbb{N}$ or $\mathbb{N^*}$. With sequences we can not talk about such things as being continuous. $\endgroup$ – Stefan Octavian Sep 26 at 15:14
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Let $\sqrt[2n+1]{-1}=y$

$y^{2n+1}=-1=e^{(2m+1)\pi i}$ where $m$ is any integer

$y=e^{(2m+1)\pi i/(2n+1)}$ where $0\le m\le 2n$

Using

Intuition behind euler's formula

$y$ will be real if $\pi$ divides $\dfrac{(2m+1)\pi}{2n+1}$

$\iff2n+1$ divides $2m+1$ which is possible if $m=n$

So, the only real value of $y$ is $-1$

Of course this is possible if $n$ remains an integer

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Let

  • $x=\frac1n\to 0^+$

  • $f(x)=(-1)^x $

  • $g(x)=\frac x{2+x}\to 0$

then the property

$$\lim_{x\to 0^+}f(g(x))=f(\lim_{x\to 0^+}g(x))$$

doesn't hold necessarly since $f(x)$ is not continuous.

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  • $\begingroup$ Please read my question carefully. I'm explicitly talking about a sequence, not a function. The limit is only for $n \in \mathbb{N}$ $\endgroup$ – Stefan Octavian Sep 26 at 14:50

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