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$\lim_{n\to\infty}h_n(x) = x\lim_{n\to\infty}x^{\frac{1}{2n-1}}$ 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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    $\begingroup$ $x^{1/(2n+1)}$ does not converge to $1$: If $x$ is negative, so is $x^{1/(2n+1)}$. That should give you a hint. Try finding what it converges to, and this should clarify the problem. $\endgroup$ – Andrés E. Caicedo Apr 24 '14 at 22:11
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We have

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

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  • $\begingroup$ You can't do that always. Can you tell me how to do it with $x^{1+\frac{1}{2n}}$? $\endgroup$ – Martigan Apr 29 '15 at 15:00
  • $\begingroup$ @sami, I'm sorry to see your account is suspended. I've only ever seen great answers from you. $\endgroup$ – Simon S Apr 29 '15 at 15:08
  • $\begingroup$ As far as I know the non-integer power of negative number is undefined, since, for example in our problem, when $x=-1$ $$(-1)^{1+\frac{1}{2n-1}}=(-1) \cdot \sqrt[2n-1]{-1}=(-1)\cdot (-1)=1$$ and $$(-1)^{1+\frac{1}{2n-1}}=(-1) \cdot (-1)^{\frac{2}{4n-2}}=(-1)\cdot \sqrt[4n-2]{(-1)^2}=-1 \cdot 1 = -1$$ So, what is a $x^{1+\frac{1}{2n-1}}$ when $x<0$ ? $\endgroup$ – Mher Apr 30 '15 at 16:28
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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.

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