A (possibly) difficult epsilon-delta proof So I'm struggling to find an epsilon-delta proof for why the function $f(h) = \dfrac{\sqrt[3]{1+h}-1}{h}$ approaches $\dfrac{1}{3}$ as $x \rightarrow 0$. 
I'd like to know how one can show using an epsilon-delta proof or some other means that for $0 < \epsilon < \dfrac{1}{8}$ and for $0 < h < 4\epsilon$, we have that $1+(\dfrac{1}{3}-\epsilon)h < \sqrt[3]{1+h}$.
I can't seem to show that for $0 < h <4\epsilon$, $\dfrac{1}{3} - \dfrac{\sqrt[3]{1+h}-1}{h} < \epsilon$.

$\lim_\limits {h\to 0} \frac{\sqrt [3]{1 + h} - 1}{h} = \frac{d}{dx} \sqrt [3] x$ evaluated at $1.$ 
First, let's write the $\epsilon - \delta$ definition of the limit 
$\forall \epsilon > 0, \exists \delta > 0 : |h| < \delta \implies |\frac{(1 + h)^\frac 13 - 1}{h} - \frac 13| < \epsilon$
Use the generalized binomial theorem to expand out the radical.
$(1 + h)^{\frac 13} = 1 + \frac 13 h - \frac {1}{9} h^2 + \frac {5}{81} h^2 - \cdots$
The series converges when $|h| < 1$ 
We can use this to find establish an upper bound and a lower bound.
$ 1 + \frac 13 h - \frac 19 h^2 \le (1 + h)^{\frac 13} \le 1 + \frac 13 h$
and so 
$- \frac 19 h\le \frac{(1 + h)^\frac 13 - 1}{h} - \frac 13 \le 0\\
|\frac{(1 + h)^\frac 13 - 1}{h} - \frac 13| < \frac {\delta}{9}$
let $\delta = \max(\frac {\epsilon}{9},1)$
