Epsilon-delta definitoon of continuity for $f : x → x^3$

(a) Let $f(x)=x^3$. Fix $x_0\in \R$  and $\epsilon >0$. Define 
\[\delta =\min \{1, \frac{\epsilon}{(1+3|x_0|(1+|x_0|))}\}.\]
For $|x-x_0|<\delta$, we have 
\[|f(x)-f(x_0)|=|x^3-x_0^3|=|(x-x_0)^3+3xx_0(x-x_0)|\]
\[=|x-x_0||(x-x_0)^2+3x x_0|\leq |x-x_0|((x-x_0)^2+|3x x_0|)\]
\[\leq |x -x_0| (\delta^2+3|x_0||x|)\leq \delta  (\delta^2+3|x_0||x-x_0+x_0|)\]
\[\leq (\delta^2+3|x_0||x-x_0|+|x_0|)\leq \delta (\delta^2+3|x_0|(\delta+|x_0|))\]
\[\leq \delta (1+3|x_0|(1+|x_0|)) \leq \frac{\epsilon}{(1+3|x_0|(1+|x_0|))} (1+3|x_0|(1+|x_0|))=\epsilon.\]
Thus $|x-x_0|<\delta$ implies 
\[|f(x)-f(x_0)| \leq \epsilon,\]
and hence $f$ is continuous at $x_0$. 

(b) Note that if $x_0\in [0,a]$. Then 
\[\frac{\epsilon}{(1+3|x_0|(1+|x_0|))} \leq \frac{\epsilon}{(1+3a(1+a))},\]
and if we define
\[\delta=\min \{1, \frac{\epsilon}{(1+3a(1+a))}\},\]
then by the above argument we conclude that 
\[|f(x)-f(x_0)| \leq \epsilon.\]
Since $\delta$ is independent of $x_0$, $f$ is uniformly continuous on $[0,a]$.