Limit of $h_n(x)=x^{1+\frac{1}{2n-1}}$ $\lim_{n\to\infty}h_n(x) = x\lim_{n\to\infty}x^{\frac{1}{2n-1}}$ where $h_n(x)=x^{1+\frac{1}{2n-1}}$. I understand that $\lim_{n\to\infty}x^{\frac{1}{2n-1}}$ goes to one but what I don't understand is how did our limit become $h_n(x)=|x|$? I'm just having hard time wrapping my head around the appearance of absolute value.
Note. This is an example (Chapter 6, Section 2) from Understanding Analysis by Abbott.
 A: We have
$$\large x^{\large 1+\frac{1}{2n-1}}=x^{\large\frac{2n}{2n-1}}=\left(x^2\right)^{\large\frac{n}{2n-1}}\xrightarrow{n\to\infty}\sqrt{x^2}=\lvert x\rvert$$
A: Key point here is that there is $2n-1$ as the denominator. This is not by accident... It would not work with $2n$...
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: The question has been asked a long time ago, but I'll give it another whirl. Roots of negative real numbers are only defined when the exponent is both fractional and when the denominator of the fraction is odd. Then we can write
$$(-x)^{\frac{a}{b}}=(-1)^{\frac{a}{b}}\cdot x^{\frac{a}{b}}=(-1)^a\cdot x^{\frac{a}{b}} \quad b \ odd$$
The $b$ vanishes because $\sqrt[b]{-1}=-1$. With this in hand we can take the limits separately for $x\geq0$ und $x<0$
$$\lim_{n\to\infty}x^{1+\frac1{2n-1}}=x\lim_{n\to\infty}x^{\frac1{2n-1}}=x \quad \text{  for  } x\geq 0$$
$$\lim_{n\to\infty}x^{1+\frac1{2n-1}}=(-1)^1x\lim_{n\to\infty}x^{\frac1{2n-1}}=-x \quad \text{  for   }  x< 0$$
Combining these results leads to our conclusion
$$\lim_{n\to\infty}x^{1+\frac1{2n-1}}=|x| \quad x\in [-1,1]$$
