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$$ h_n (x) = x^{1 + {1 \over 2n - 1}}$$ for $x \in [-1,1]$.

Why is $\lim_{n \to \infty} h_n(x) = |x|$? I know that for $C > 0$ it holds that $$ \lim_{n \to \infty} C^{1\over n} = 1$$

but I don't know why the other limit should hold. Especially not $$ \lim_{n \to \infty} C^{1\over n} = -1$$ if $C < 0$.

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up vote 1 down vote accepted

Note that $h_{n}(x)=x\cdot x^{1/(2n-1)}$, and since $n$ is a natural number, this is an odd root of $x$. In particular, it preserves the sign, since $$(-x)^{1/r}=-x^{1/r}$$ is true iff $r$ is odd (for a proof, raise both sides to the power $r$).
So when $x\ge0$ you have seen the result is obviously true, and if $x<0$ we can apply the above result.

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Wow thank you, it's so obvious now. – blue May 2 '14 at 13:12

One can use the convention that $x^{p/q}$ ($p$ and $q$ integers, $q\ne0$) is well defined for any $x$ if the denominator $q$ is odd.

Since $$ 1+\frac{1}{2n-1}=\frac{2n}{2n-1} $$ and $n$ is assumed to be a positive integer, we are in that situation. Moreover, the numerator is even and so $$ x^{2n/(2n-1)}=|x|^{2n/(2n-1)}. $$ Since $$ \lim_{n\to\infty}\frac{2n}{2n-1}=1 $$ you're done, because, for $a>0$, the function $t\mapsto a^t$ (defined for any real $t$) is continuous.

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