Hey everyone. I am having trouble understanding the following statements from my textbook (Essential Calculus by Stelwart)

"..We showed that iffis differentiable ata, then

$\displaystyle \Delta y= f'(a)\Delta x + \epsilon\Delta x$ where $\displaystyle \epsilon \rightarrow 0$ as $\displaystyle \Delta x \rightarrow 0$"

What does the epsilon delta x have to do with this equation? I divided everything by delta x and got this:

$\displaystyle \frac{\Delta y}{\Delta x} - f'(a) = \epsilon $

I'm assuming lim is involved so this is what it would be:

$\displaystyle \lim_{x\rightarrow 0}\frac{\Delta y}{\Delta x} - f'(a) = \epsilon $

So is epsilon error? If so, why would there be error? I'm definitely confused.

The book goes on to say more:

Now consider a function of two variables, z = f(x,y), and the suppose x changes from a to $\displaystyle a+\Delta x$ and y changes from b to $\displaystyle b+\Delta y$. Then the correspondingincrementof z is

$\displaystyle \Delta z = f(a + \Delta x,b + \Delta y) - f(a,b)$

Thus, the increment delta z represents the change in the value of f when (x,y) changes from (a,b) to $\displaystyle (a+\Delta x, b+\Delta y)$. by analogy with (5) (which is the first equation with the epsilon crap) we define the differentiability of a function of two variables as follows:

If z = f(x,y), then f is differentiable at (a,b) if delta x can be expressed in the form

$\displaystyle \Delta z = f_x(a,b)\Delta x + f_y(a,b)\Delta y + \epsilon_1\Delta x + \epsilon_2\Delta y$

where $\displaystyle \epsilon_1$ and $\displaystyle \epsilon_2 \rightarrow 0$ as $\displaystyle (\Delta x,\Delta y) \rightarrow (0,0)$.

It makes sense except the epsilon stuff.

Any help is appreciated and thanks in advance!