Let $f$ be a function with domain $R$ that satisfies the conditions: $$f(x+y)=f(x)f(y), \forall x,y $$ and $$f(0) \neq 0$$
(a) Show that $f(0)=1$
(b) Prove that $f(x) \neq 0$, for all $x\in R$
(c) Assuming that $f'(x)$ exists for all $x \in R$, use the definition of the derivative to show that $f(x)$ satisfies the equation $f'(x)=kf(x)$, where $k=f'(0)$
I've tried solving part (a) and (b) by substituting $x=y=0$ and $y=-x$ respectively, but I can't seem to solve part (c) as I can't avoid dividing by 0 when dealing with the limit. Does anyone know how to work around this?