# Thread: Slope by limit process

1. ## Slope by limit process

Greetings,

I need to use the limit definition of :

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

( / means all divided by $\Delta x$)

I have to use the limit definition to find the slope of the tangent line to the graph of $f$ given at these equations:

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

If someone could show me how to solve this step by step that would be cool.

2. $f(x)=4-x^2$
$f'(x)=-2x$

$m=f'(x)=-4$

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

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

Now as you said:

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

We can do some cancelling
$m=\lim_{\Delta x\rightarrow 0} \frac{-4\Delta x - (\Delta x)^2}{\Delta x}$

We can do some more cancelling which leaves us with
$m=\lim_{\Delta x\rightarrow 0} -4 - \Delta x$

The limit is as $\Delta x \rightarrow 0$, so sub it in, which gives
$m= -4-0
=-4$

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

You can also verify this by differentiating $f(x)$ as alexmahone did:
$f' (x)=-2x$
$f'(2)=-2(2) = -4$

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