Why is the real function $x^{1/3}$ continuous?

I want to prove that the function $f(x) := x^3$ for all real $x$ defines a homeomorphism from $\mathbb{R}$ to $\mathbb{R}$. But I am finding it difficult to prove that the inverse map is continuous!

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Suppose $x^{1/3}$ is not continuous at some $x_0\in\mathbb R$, so for some $\epsilon>0$ we have a sequence $(x_n)$ such that $\lim\limits_{n\to\infty} x_n = x_0$ yet $|x_n^{1/3}-x_0^{1/3}|>\epsilon$ for all $n$. We must have either infinitely many $n$ such that $x_n^{1/3}-x_0^{1/3}$ or infinitely many $n$ such that it is negative. Similarly, either infinitely many of these $x_n$ are $\geq 0$ or infinitely many are $\leq 0$. By relabeling our $x_n$ we can assume that neither of these signs change for any $n$. I will deal with the case $x_n\geq 0$ for all $n$ and leave the other to you. If $x_n^{1/3}-x_0^{1/3}$ is positive for all $n$, then $x_n^{1/3}>x_0^{1/3}+\epsilon$ so $x_n>x_0+3x_0^{2/3}\epsilon+3x_0^{1/3}\epsilon^2+\epsilon^3\geq x_0+\epsilon^3$ for all $n$, contradicting the fact that $\lim\limits_{n\to\infty} x_n=x_0$. If it is negative for all $n$, then $x_0^{1/3}>x_n^{1/3}+\epsilon$ so $x_0>x_n+3x_n^{2/3}\epsilon+3x_n^{1/3}\epsilon^2+\epsilon^3\geq x_n+\epsilon^3$ for all $n$, again contradicting the fact that $\lim\limits_{n\to\infty} x_n=x_0$. Thus $x^{1/3}$ is continuous at any $x_0\in\mathbb R$, so $x^{1/3}$ is continuous.
This is more than a hint, as the proof is trivial outside $[-1,1]$. –  leftaroundabout Feb 19 '12 at 13:02