Let $\displaystyle f(x) = x^3$.

1.) Prove (using the $\displaystyle \epsilon-\delta$ definition) that at $\displaystyle x = 2, f$ is continuous on $\displaystyle \mathbb{R}$.

2.) Prove (using the def. of differentiability) that at $\displaystyle x = 4, f$ is differentiable.

Well, I don't know if this is helpful, but I know $\displaystyle f$ is not uniformly continuous on $\displaystyle \mathbb{R}$. There is a similar example in my book proving that x^2 is continuous on $\displaystyle \mathbb{R}$. They do:

Given $\displaystyle c \in \mathbb{R}$:

$\displaystyle |f(x) - f(c)| = |x^2 - c^2| = |x-c||x+c|$

Now they find an upperbound on $\displaystyle |x+c|$ by makings ure $\displaystyle \delta$ does not exceed 1. Apparently this insures all values of $\displaystyle x$ will fall in the interval $\displaystyle (c-1,c+1)$.

I don't entirely get that ste.

Then, they get:

$\displaystyle |x + c| \leq |x| + |c| \leq (|c|+1) + |c| = 2|c| + 1$

They let $\displaystyle \epsilon > 0$ be given, then choose $\displaystyle \delta = min\left(1,\frac{\epsilon}{2|c|+1}\right)$, then $\displaystyle |x - c| < \delta$

$\displaystyle \implies |f(x) - f(c)| = |x-c||x+c|< \frac{\epsilon}{2|c|+1}(2|c|+1) = \epsilon$.

I'm assuming it's similar for $\displaystyle x^3$?