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Define $h_n(x)=x^{1+\frac{1}{2n-1}}$ on the domain $[-1,1]$. Then, according to my real analysis book, the pointwise limit of $h_n(x)$ is $|x|$.

I tried to look at this limit informally (calc 1 style, in a sense) and I figured that the limit on this domain would be $x$. This was due to separating the function into $x\cdot x^{\frac{1}{2n-1}}$ and claiming that the second term in the product approaches $1$ as $n$ approaches infinity. Perhaps there is some fundamental flaw in my thinking. I'd like to know how the absolute value function is the pointwise limit and so any help is appreciated.

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    I don't think that this is true. I think it does converge to $x$ (though not because of what you describe). What book are you using? – 2017-01-10
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    This is of course awfully sloppy work from your book but my guess is that they define $x^{1/(2n-1)}$ in the canonical way if $x\geqslant0$ and as $\mathrm{sign}(x)\cdot|x|^{1/(2n-1)}$ if $x<0$. This definition ensures that $x^{1/(2n-1)}=w$ is such that $w^{2n-1}=x$ for every $x$, as desired. Then indeed $x^{1/(2n-1)}\to\mathrm{sign}(x)$ for every $x\ne0$ and $x\cdot x^{1/(2n-1)}\to|x|$ for every $x$ when $n\to\infty$. – 2017-01-10
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    @Did who defines it as so? Please have a look at this: http://www.wolframalpha.com/input/?i=lim(n+-%3E+infinity)+(-1%2F2)%5E(2n%2F(2n-1)), for instance. – 2017-01-10
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    @Did by the way, that my answer doesn't match the user's textbook's expectations doesn't warrant a downvote. – 2017-01-10
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    @OpenBall Using WA to settle a question of definition is of course horribly misguided. Just for kicks, try entering [this](http://www.wolframalpha.com/input/?i=(-1%2F2)%5E(4%2F3)) into the beast... – 2017-01-10
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    @Did the point is that your definition is not universal. – 2017-01-10
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    @OpenBall "who defines it as so?" Dunno, this is simply, imho, the reason why the OP's book suggests the limit $|x|$. – 2017-01-10
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    @Did and now you suggest its deletion?! Seriously, people can have different opinions. Downvotes are "socially acceptable", as they in large reflect the diversity of opinions/conventions, but it is quite absurd that a disagreement over a definition might prompt a request for the deletion of a post. – 2017-01-10
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    @OpenBall I thought you were also putting the downvote in the basket of the capital sins... Not anymore? Anyway, since you have no way of knowing who votes what, your hypotheses on this topic are unwarranted and I suggest that you concentrate on some more fascinating subjects. If you want to get some input on the reasons why answers can be downvoted and/or voted for deletion, I am sure you will find some page on the site explaining these in details. – 2017-01-10
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    @Did downvotes for me are not in the basket of capital sins, as is evident from my own downvoting history. I was merely trying to understand your point of view, and I did. Please quit this patronizing manner of speaking, as it is offensive; "I suggest you concentrate on some more fascinating subjects" - this is ad hominem nonsense. – 2017-01-10
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    @OpenBall Do you know the meaning of "ad hominem"? Anyway, kudos to you for having derailed this thread from the mathematical question at hand. – 2017-01-10
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    @Did I hope you do. – 2017-01-10
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    Sorry for the delay. I had to step away for a little while. The book I'm using is the first edition of Abbott's "Understanding Analysis". Im beginning to think that there may be a misprint in the definition of $h_n(x)$. – 2017-01-10
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    No misprint, see my explanation (or the answer you accepted...). – 2017-01-10
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    @Did thank you for your help. I see why this is the case now. – 2017-01-10

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For negative $x$, the odd degree roots are also negative, if defined at all. Thus in the product you get a positive quantity. You can also rewrite this expression as $$ \left(x^2\right)^{\frac{n}{2n-1}} $$ to get a clearer convergence towards the absolute value.