proof of derivation product rule

I have question regarding to product rule.

We know that derivative of y with respect to x is:

$\displaystyle

\frac{dy}{dx} = f'(x) = \lim_{h\rightarrow 0} \: \frac{f(x+h) - f(x)}{x+h-x} = \lim_{h\rightarrow 0} \: \frac{f(x+h) - f(x)}{h} = \frac{\Delta y}{\Delta x}$

Now I'm trying to understand the proof of product rule:

$\displaystyle f'(x)\cdot g'(x) = f'(x)\cdot g(x) + f(x)\cdot g'(x)

$

to simplify notation:

$\displaystyle let\;u=f(x)\;and\;v=g(x).\;\;Then \;D(u\cdot v) = u' \cdot v + u \cdot v'$

Ok, now to proof:

$\displaystyle (u \cdot v)' = \lim_{h\rightarrow 0} \frac{\left[u(x+h) \cdot v(x+h) \right] - \left[u(x) \cdot v(x) \right]}{h}

$

we know that $\displaystyle \Delta u = u(x+h) - u(x) \;,\; thus\; u(x+h) = \Delta u + u(x) $ so we can substitute this with respected to u and v in equation and get:

$\displaystyle (u \cdot v)' = \lim_{h\rightarrow 0} \frac{\left[(\Delta u+u(x)) \cdot (\Delta v + v(x)) \right] - \left[u(x) \cdot v(x) \right]}{h}$

Now to do some algebra we get:

$\displaystyle = \lim_{h\rightarrow 0} \left\{ \frac{\Delta u \cdot \Delta v}{h} + \frac{\Delta u}{h} \cdot v(x) + \frac{\Delta v}{h} \cdot u(x) + \frac{u(x) \cdot v(x)}{h} - \frac{u(x) \cdot v(x)}{h} \right\} $

$\displaystyle = \lim_{h\rightarrow 0} \left\{ \frac{\Delta u \cdot \Delta v}{h} + \frac{\Delta u}{h} \cdot v(x) + \frac{\Delta v}{h} \cdot u(x) \right\}$

And we know that $\displaystyle u' = \lim_{h\rightarrow 0} \frac{\Delta u}{h} \; , \; v' = \lim_{h\rightarrow 0} \frac{\Delta v}{h}$

so finally we substitute u' and v' to the equation we get:

$\displaystyle = \lim_{h\rightarrow 0} \frac{\Delta u \cdot \Delta v}{h} + u' \cdot v + v' \cdot u $

How i can get rid of $\displaystyle \lim_{h\rightarrow 0} \frac{\Delta u \cdot \Delta v}{h}$ since this is some error in my proof.