I think what he needs is to solve this using First Principle.

$\displaystyle f(x) = 4-x^2$ at point $\displaystyle (2,0)$, so $\displaystyle x=2$

Now as you said:

$\displaystyle m= \lim_{\Delta x\rightarrow 0} \frac{f(x+\Delta x)- f(x)}{\Delta x}$

Sub $\displaystyle x=2$

$\displaystyle m=\lim_{\Delta x\rightarrow 0} \frac{f(2+\Delta x)- f(2)}{\Delta x}$

$\displaystyle m=\lim_{\Delta x\rightarrow 0} \frac{4-(2+\Delta x)^2- (4-(2)^2)}{\Delta x}$

$\displaystyle m=\lim_{\Delta x\rightarrow 0} \frac{4-(2^2+4\Delta x + (\Delta x)^2)- (4-(2)^2)}{\Delta x}$

$\displaystyle m=\lim_{\Delta x\rightarrow 0} \frac{4-4-4\Delta x - (\Delta x)^2- 4+4}{\Delta x}$

We can do some cancelling

$\displaystyle m=\lim_{\Delta x\rightarrow 0} \frac{-4\Delta x - (\Delta x)^2}{\Delta x}$

We can do some more cancelling which leaves us with

$\displaystyle m=\lim_{\Delta x\rightarrow 0} -4 - \Delta x$

The limit is as $\displaystyle \Delta x \rightarrow 0$, so sub it in, which gives

$\displaystyle m= -4-0

=-4$

So, at the point $\displaystyle (2,0)$, $\displaystyle m=-4$

You can also verify this by differentiating $\displaystyle f(x)$ as alexmahone did:

$\displaystyle f' (x)=-2x$

$\displaystyle f'(2)=-2(2) = -4$

Which is the same as what we obtained before, although this doesn't directly use the first principles or limits.