Derivative Product Rule Proof

If we let p(x) = f(x)g(x) then using the limit definition, we do: $p^{\prime }(x)={\lim }_{h\to 0}[\frac{f(x+h)g(x+h)-f(x)g(x)}{h}]$ $p^{\prime }(x)={\lim }_{h\to 0}[\frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}]$ $p^{\prime }(x)={\lim }_{h\to 0}[\frac{f(x+h)-f(x)}{h}g(x)+f(x)\frac{g(h+x)-g(x)}{h}]$ $p^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}{\lim }_{h\to 0}g(x)+{\lim }_{h\to 0}f(x){\lim }_{h\to 0}\frac{g(h+x)-g(x)}{h}$ $p^{\prime }(x)=f^{\prime }(x)g(x)+g^{\prime }(x)f(x)$