Uniform Continuity of $f(x)=x^3$ 1.)Determine whether $f(x)=x^3$ is uniformly continuous on [0,2)
So far, I have $\delta$ = 2 and $\epsilon$ = 8, and plan on using the sandwich theorem with $x^2$ and eventually equating $\delta = \epsilon$. 
 A: For given $\epsilon >0$, choose $\delta=\epsilon/12$, then for $x,y\in [0,2),$ such that 
$|x-y|<\delta\implies |f(x)-f(y)|=|x^3-y^3|$
$=|x-y||x^2+xy+y^2|<|x-y|(|x|^2+|x||y|+|y|^2)<12|x-y|=12\delta=\epsilon$
Thus, for $\epsilon>0$, $\exists \delta >0$(independent of point where continuity is to be checked), such that $|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$.
Hence, Proved. 
A: For a quick way, recall that any continuous function is uniformly continuous on a closed interval $[a,b]$. If $f(x)$ is uniformly continuous on $[a,b]$ then it must be uniformly continuous on $[a,b)$ (with the same $\delta$ for every $\epsilon$).
A: A related problem.
Hint: Another approach is the mean value theorem. Here is a related problem.
Added: 

Mean Value Theorem: If a function $f$ is continuous on the closed interval [a, b], where a < b, and differentiable on the open interval (a, b), then there exists a point c in (a, b) such that
  $$ f'(c) = \frac{f(b) - f(a)}{b-a}. $$

In your case, we start as 
$$ \frac{f(x)-f(y)}{x-y}=f'(\eta),\,\,\eta\in(x,y) \implies \frac{x^3-y^3}{x-y}=3\eta^2 $$
$$ \implies |{x^3-y^3}|=|3\eta^2||x-y|\leq 12|x-y| < \epsilon $$ 
$$\implies |x-y| < \frac{\epsilon}{12}=\delta. $$
Now, choosing $ \delta=\frac{\epsilon}{12} $, we have

$$ |x-y|< \delta \implies |x^3-y^3|<\epsilon.  $$

