Use the definition to find the derivative of $g(x)=x^3$. Use the definition to find the derivative of $g(x)=x^3$.
        We consider 
            $\lim_{x\to x_0}\frac{x^3-x_0^3}{x-x_0}$
        where $x_0$ is an accumulation point of the domain $D$ and $x_0 \in D$. We choose $\delta$ such that $0<|x-x_0|<\delta$ gives $|\frac{x^3-x_0^3}{x-x_0}|<\varepsilon$. We examine the following guessing that our limit is $3x^2$??
I am in analysis class so this isn't as simple as this problem at first seems though, it is very simply and I'm just having a brain malfunction. Thanks for the help!
 A: If $x_{o} \in D$, then
$$
\frac{g(x) - g(x_{o})}{x-x_{o}} = \frac{x^{3} - x_{o}^{3}}{x-x_{o}} = x^{2}+xx_{o} + x_{o}^{2} \to x_{o}^{2} + x_{o}^{2} + x_{o}^{2} = 3x_{o}^{2}
$$
as $x \to x_{o}$;
to prove that $f'(x_{o}) = 3x_{o}^{2}$ we have to appeal to the definition of limit that you write in your question.
A: Just to be very formal about this one I wanted to post an answer. Feel free to comment if it has any errors. 
We begin with $\lim\limits_{x \to x_0}\frac{x^3-x_0^3}{x-x_0}$. From here we do a little simplifying so we may effectively guess at a limit $\lim\limits_{x\to x_0}x^2+xx_0+x_0^2$ so we guess our limit will tend to $3x_0^2$. From there we wish to find a $\delta$ such that $0<|x-x_0|<\delta$ implies we have $|x^2+xx_0+x_0^2-3x_0^2|< \varepsilon$. 
We have
$|x^2+xx_0+x_0^2-3x_0^2|=|x^2+xx_0-2x_0^2|=|x^2+2xx_0-xx_0-2x_0^2|=|(x-x_0)(x+2x_0)|<|\delta(x+2x_0)|=|\delta(x-x_0+3x_0)|\leq|\delta|(|(x-x_0)|+|3x_0|$
and so we restrict $\delta<1$ giving $|\delta|(1+|3x_0|)<\varepsilon$ and so $\delta < \frac{\varepsilon}{1+|3x_0|}$ therefore we select $\delta=min\{1,\frac{\varepsilon}{1+|3x_0|}\}$ and we are done.
