# Derivation af oa product

1. Nov 13, 2011

### georg gill

What do I do wrong here:

$$\frac{f(x+h)g(x+h)-f(x)g(x)}{x+h-x}=\frac{f(x+h)g(x+h)}{h}-\frac{f(x)g(x)}{h}=\frac{f(x+h)}{h}g(x+h)-\frac{f(x)}{h}g(x)=\lim_{h \to 0}\frac{f(x+h)}{h}\lim_{h \to 0}g(x+h)-\lim_{h \to 0}\frac{f(x)}{h}g(x)$$

$$\lim_{h \to 0}g(x+h)=g(x)$$

$$\lim_{h \to 0}\frac{f(x+h)}{h}\lim_{h \to 0}g(x+h)-\lim_{h \to 0}\frac{f(x)}{h}g(x)=\lim_{h \to 0}\frac{f(x+h)}{h}g(x)-\lim_{h \to 0}\frac{f(x)}{h}g(x)=(\lim_{h \to 0}\frac{f(x+h)}{h}-\lim_{h \to 0}\frac{f(x)}{h})g(x)=(\lim_{h \to 0}\frac{f(x+h)-f(x)}{h})g(x)=\frac{df}{dx}g(x)$$

I know how they derieve the derivation of a product:

http://bildr.no/view/918745

But how come what I did above does not work?

Last edited: Nov 13, 2011
2. Nov 13, 2011

### micromass

You did something like

$$\lim_{x\rightarrow a}{f(x)g(x)+h(x)}=\lim_{x\rightarrow a}{f(x)}\lim_{x\rightarrow a}{g(x)}+\lim_{x\rightarrow a}{h(x)}$$

But this is not true. You can't do that.

It is ONLY true if $\lim_{x\rightarrow a}{f(x)}$, $\lim_{x\rightarrow a}{g(x)}$ AND $\lim_{x\rightarrow a}{h(x)}$ converge. In your example, you don't have that. For example

$$\lim_{h\rightarrow 0}{\frac{f(x)}{h}}$$

does not converge.