Why does $\lim_{n \to \infty} x^{1+1/(2n-1)}=|x|$? Consider the following limit for $x \in [-1,1]$ and $n \in \mathbb N$:
$$\lim_{n \to \infty} x^{1+\frac{1}{2n-1}}=|x|$$
I can't see where does the absolute value come from. Why isn't the answer simply $x$?

This limit is available on page 156 of Abbott's Understanding Analysis.

EDIT: Sorry i found a dublicate here.
 A: This can also be done this way :
$$\lim_{n\to\infty}x^{1+ \frac{1}{2n-1}}= \lim_{n\to\infty}x^{\frac{2n-1+1}{2n-1}}= \lim_{n\to\infty}x^{\frac{2n}{2n-1}}$$
$$= \lim\limits_{n\to\infty}\left(x^2\right)^{\frac{n}{2n-1}}= \left(x^2\right)^{ \lim\limits_{n\to\infty} \frac{n}{2n-1}}$$
$$= \left(x^2\right)^{ \lim\limits_{n\to\infty} \frac{1}{2-\frac1n}}= \left(x^2\right)^{\frac{1}{2-0}} = \left(x^2\right)^{\frac{1}{2}}=\sqrt{x^2}$$
Since, $x\in [-1,1]$
$$=|x|$$
Hence proved
A: EDIT: key point here is that there is $2n-1$ as the denominator. This is not by accident...
Let's call $y_n=x^{\frac{1}{2n-1}}$
$y_n^{2n-1}=x$ has the same sign as $y_n$, since $2n-1$ is even. 
For $x<0$, $y_n^{2n-1}=-|x|$,   $\forall n \in \mathbb{N}$
$y_n$ can be written as (since it has the same sign as $y_n^{2n-1}):
$y_n=-|x|^{\frac{1}{2n-1}}$ 
$\lim_{n \to \infty} y_n=-1$
Thus
$\lim_{n \to \infty} |x|^{1+\frac{1}{2n-1}}=-x=|x|$
This solves the case $x<0$. The other case is obvious. 
A: |x| implies x < 0 0r x > 0, i.e. x has a possibility of being either. 1/(2n - 1) approaches 0 as n -> infinity and therefore the limit is x^1 since x^0 = 1, and then it follows that if x < 0, this limit is negative and positive is x > 0. All it means is that x can be positive or negative.
