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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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  • $\begingroup$ 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? $\endgroup$ – user384138 Jan 10 '17 at 14:25
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    $\begingroup$ 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 $n\to\infty$. $\endgroup$ – Did Jan 10 '17 at 14:34
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    $\begingroup$ @OpenBall Using WA to settle a question of definition is of course horribly misguided. Just for kicks, try entering this into the beast... $\endgroup$ – Did Jan 10 '17 at 14:50
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    $\begingroup$ @OpenBall "who defines it as so?" Dunno, this is simply, imho, the reason why the OP's book suggests the limit $|x|$. $\endgroup$ – Did Jan 10 '17 at 14:51
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    $\begingroup$ No misprint, see my explanation (or the answer you accepted...). $\endgroup$ – Did Jan 10 '17 at 16:43
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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.

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