The derivative is a real limit? In my classes, the derivative is usually defined as "the limit of the fractional incremental ratio". But I found out another way to define the derivative from an old book from Gardner & Thompson's "Calculus Made Easy". For example, if we have $f(x) = x^2$ and we want to calculate the derivative:
$$
f(x) = x^2
$$
So our y is
$$
y = x^2
$$
Right now we are only considering the increment of y and x so we can calculate that this way:
$$
y+dy = (x+dx)^2
$$
$$
y+dy = x^2+dx^2+2xdx
$$
I can remove
$$
dx^2
$$
because it's a very small quantity related to our magnitude.
The result is
$$
y+dy = x^2+2xdx
$$
I subtract the original quantity
$$
y+dy-y=x^2+2xdx-x^2
$$
$$
dy=2xdx
$$
$$
dy/dx=2x
$$
The derivative is equal to $$2x$$ and I calculate that without using any limits. So, my question is: the derivative is a real limit? What about the orders of magnitude? A limit is the representation of a value related to our magnitude?
 A: That's exactly what limits are about: what happens when h (or dx, or whatever you call it) is very small.
More precisely, recall that $\lim f(h)=a$ iff $f(h)=a+o(1)$ (little-o notation); so $\lim\frac{f(x+h)-f(x)}h=a$ iff $\frac{f(x+h)-f(x)}h=a+o(1)$ or equivalently iff $f(x+h)=f(x)+ah+o(h)$. The last statement is just more formal way to say that $f(x+h)$ is equal to $f(x)+ah$ "up to next orders of magnitude".
This point of view of yours is, actually, not only more natural but also easier to generalize (in the direction of, say, multivariable calculus).
A: You actually are doing exactly the limiting process, only it's a bit hidden.  If you keep that $dx^2$ term for a bit longer, we get 
$$ dy = 2x dx + dx^2 $$
and so
$$ \frac{dy}{dx} = 2x + dx. $$
Taking "$dx^2$ to be small" amounts to setting $dx=0$ on the right hand side of that last equation.  Now let's compare this to the usual limiting method:
$$ \lim_{h\to 0} \frac{(x+h)^2-x^2}{h}$$
$$ = \lim_{h\to 0} \frac{x^2+2hx+h^2-x^2}{h}$$
$$ = \lim_{h\to 0} \frac{2hx+h^2}{h} $$
$$ = \lim_{h\to 0} (2x+h).$$
In that last expression we can plug in $h=0$ to get that the derivative of $x^2$ is $2x$.
We get to the exact same expression (except instead of a $dx$ we have an $h$) in which we set our little increment to $0$.
In general, many limits can be evaluated by simplifiying algebraically and then just plugging in the point you're looking at.  When you are taking the limit as something goes to zero, lots of times the "plugging in" part boils down to "this gets really small so let's ignore it."
