Slope by limit process

**AMaccy** · May 29th 2009, 08:01 PM

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.

**alexmahone** · May 29th 2009, 09:00 PM

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

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

**Enedrox** · May 29th 2009, 09:12 PM

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
$m= -4-0<br /> =-4$

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.

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