Wondering if my reasoning is correct for a tricky calculus limit word problem and thanks for reading my post.  The problem reads like this:
If a book starts to fall off a building from 120 feet in the air and the position of the book after t seconds is given by s(t) = -16t2) + 120.  Find the velocity of the book at t=2 seconds by calculating the limit of the velocity:
velocity = lim as x -> 0 of s(2 + x) - s(2) / x
Since we know that t = 2, we can find the position at that time by plugging in 2 to the equation to get -16(4)+120 = 64.    We plug that into the velocity function where s(2) is:


*

*s(2 + x) - 56 / x =

*(-16 (2 + x)2 + 120 - 56 / x = 

*-16x2-64x-64+64 =

*-16x - 64


So as x goes to 0, the velocity becomes -64 feet per second (downward).  I'm not sure if this is correct but it does seem reasonable.  What confuses me here is the velocity function: the s(2 + x), what is it referring to?  Since s(2) is referring to the position at time = 2, then 2 + x must refer to a position greater than 2, right?  Sort of like the slope equation for a line?
 A: The expression "Find the velocity of the book at $t=2$ seconds by calculating the limit of the velocity" is very missleading. What you really need to calcuate (and what you indeed did calculate) is the derivative of the position, which happens to be the velocity.
Say the book is at position $s(t)$ at time $t$, and let $\Delta$ be some time difference. Then the distance that the book travels (well, falls) between time $t$ and time $t+\Delta$ is $$
  s(t + \Delta) - s(t).
$$
The average speed, or velocity, of the book between the time $t$ and the time $t+\Delta$ is thus $$
  \frac{s(t+\Delta) - s(t)}{\Delta},
$$
because velocity is simply the distance travelled divided by the time it took.
To get the speed as a particular point in time, as opposed to the average speed within an interval of time, you find the limit of that expression as the length of the interval, i.e. $\Delta$, goes to zero. Therefore, the speed or velocity at time $t$ is $$
  v(t) = \lim_{\Delta \to 0} \frac{s(t+\Delta) - s(t)}{\Delta}.
$$
Which, of course, is simply the derivative of $s$. Your assessment that this is "something like the slope equation" is therefore spot-on! It's exactly the slope equation, but since we're interested in the "slope" at a particular point, we have to let $\Delta$ go to zero. Otherwise we'd get the average slope over the interval $[t,t+\Delta]$.
In your example, $t=2$, and you correctly computed (I'm writing $\Delta$ instead of $x$) $$\begin{eqnarray}
  \lim_{\Delta \to 0}\frac{s(2+\Delta) - s(2)}{\Delta}
  &=& \lim_{\Delta \to 0}\frac{-16(2+\Delta)^2 + 120 + 16\cdot 2^2 - 120}{\Delta} \\
  &=& \lim_{\Delta \to 0}\frac{-16\cdot 4 \cdot \Delta - 16\Delta^2}{\Delta}
  = -64.
\end{eqnarray}$$
So $s(2+\Delta)$ is the position of the book "shorty after" time $2$ (to be precise, $\Delta$ units after $2$), $s(2)$ is the position at time $2$. The difference is the amount the book travelled within that "very short" interval between times $2$ and $2+\Delta$, and the quotient is hence the average speed within that "very short" interval. The limit as the interval length goes to zero is the speed at time $2$.
