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.

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We have

$$\large x^{\large 1+\frac{1}{2n-1}}=x^{\large\frac{2n}{2n-1}}=\left(x^2\right)^{\large\frac{n}{2n-1}}\xrightarrow{n\to\infty}\sqrt{x^2}=\lvert x\rvert$$

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}):


$\lim_{n \to \infty} y_n=-1$


$\lim_{n \to \infty} |x|^{1+\frac{1}{2n-1}}=-x=|x|$

This solves the case $x<0$. The other case is obvious.