Use continuity to show that $f(x)=x^3$ is uniformly continuous on $[0,1]$ but not $[0,\infty]$ I'm trying to use continuity to show that $f(x)=x^3$ is uniformly continuous on $[0,1]$ but not $[0,\infty)$.
I've tried setting up an epsilon-delta proof, but I'm struggling a little:
By definition of uniform continuity, we know that $\forall \epsilon >0, \exists \delta >0$ such that
$|x-y|<\delta \Rightarrow |f(x) - f(y)| < \epsilon$.
And so, forcing  
$\delta = \min\{1, \frac{\epsilon}{p_x}\}$ where $p_x = (x^2+xy+y^2)$
And so, we havve that 
$|(x)^3 - (y)^3|=|(x-y)(x^2+xy+y^2)| < |\delta (x^2 + xy+y^2)| < \epsilon$
I'm not sure if this is the correct way to go about proving it, or if I landed myself into a circular argument. Furthermore, intuitively I'm guessing we only have uniform continuity on $[0,1]$ but not [0,$\infty)$ because our $p_x$ would get too large?

To show that $x^3$ fails to be uniformly continuous on $[0,\infty)$, we take $\epsilon=\frac{3}{2}$.  Then, for all $\delta>0$, and for $x=\frac{1}{\sqrt\delta}$ and $y=\frac{1}{\sqrt \delta}+\frac{\delta}{2}$ we have $|x-y|<\delta$ and
$$\begin{align}|x^3-y^3|&=\left|\left(\frac{1}{\sqrt\delta}+\frac{\delta}{2}\right)^3-\left(\frac{1}{\sqrt\delta}\right)^3\right|\\\\
&\ge \frac32\\\\
&=\epsilon
\end{align}$$
