# Thread: Slope by limit process

1. ## Slope by limit process

Greetings,

I need to use the limit definition of :

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

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

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

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

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

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

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

3. 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.