$\lim\limits_{h\to0} \frac{(x+h)^3 -x^3}{h}$ Compute the following limit I can not seem to figure out the direction to go with this problem. I'm not quite sure how to break it up.

Others have pointed out here that $(x+h)^3 = x^3+3x^2h+3xh^2+h^3$.
Here's different approach.  Recall that $a^3-b^3=(a-b)(a^2+ab+b^2)$.  Consequently
\begin{align}
(x+h)^3-x^3 & = \Big((x+h)-x\Big)\Big( (x+h)^2 +(x+h)x + x^2 \Big) \\[12pt]
& = h\Big( (x+h)^2 +(x+h)x + x^2 \Big).
\end{align}
Now we might be tempted to do the routine simplifications of the expression in the large parentheses, but it's quicker to do this:
$$
\frac{(x+h)^3-x^3} h = \frac{h\Big( (x+h)^2 +(x+h)x + x^2 \Big)} h = (x+h)^2 +(x+h)x + x^2.
$$
We want the limit of that as $h\to0$, and now we can just plug in $h=0$ and get $x^2+x^2+x^2$.
