What is $\lim_{n\to\infty} \root{2n+1} \of {-1} ?$ 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.
 A: 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.
A: 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 $$\lim_{n\to\infty} \root{2n+1}\of{-1}=-1 $$ 
A: 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.
A: 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.)
A: 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
A: 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.
