An explanation of the pointwise limit of $x^{1+1/(2n-1)}$ when $n\to\infty$ 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.

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.
