Here's the proof, as provided by text:

$\displaystyle \frac{d}{dx}(uv)=\lim_{h\to\0}\frac{u(x+h)v(x+h)-u(x+h)v(x)+u(x+h)v(x)-u(x)v(x)}{h}$$\displaystyle =\lim_{h\to\0}[u(x+h)\frac{v(x+h)-v(x)}{h}+v(x)\frac{u(x+h)-u(x)}{h}]$

$\displaystyle =\lim_{h\to\0}u(x+h)*\lim_{h\to\0}\frac{v(x+h)-v(x)}{h}+v(x)*\lim_{h\to\0}\frac{u(x+h)-u(x)}{h}.$

I understand that $\displaystyle u(v+h)v(x)-u(v+h)v(x)$ is introduced into the expression so that the equation can be expressed by definition as $\displaystyle \frac{dv}{dx}$ and $\displaystyle \frac{du}{dx}$. Although in step 2, wouldn't the denominator, h, retain its form into step 3 as:

$\displaystyle \lim_{h\to\0}\frac{u(x+h)}{h}*\lim_{h\to\0}\frac{v (x+h)-v(x)}{h}+\lim_{h\to\0}\frac{v(x)}{h}*\lim_{h\to\0} \frac{u(x+h)-u(x)}{h} ?$

It seems to me the h was omitted from a factor for each term.