Let $$f : R \to R$$ satisfy the equation $$f(x+y) = f(x) \cdot f(y)$$ for all $$x, y \in R$$ and $$f(x) \neq 0$$ for any $$x \in R$$. If the function $$f$$ is differentiable at $$x = 0$$ and $$f'(0) = 3$$, then $$\lim_{h \to 0} \frac{1}{h}(f(h) - 1)$$ is equal to ___.

The functional equation $$f(x+y) = f(x) \cdot f(y)$$ with $$f(x) \neq 0$$ for all $$x$$ tells us that $$f$$ is an exponential function. Setting $$x = y = 0$$ gives $$f(0) = f(0)^2$$, so $$f(0) = 1$$ (since $$f(0) \neq 0$$).

Now consider the limit $$\lim_{h \to 0}\dfrac{f(h) - 1}{h}$$. Since $$f$$ is differentiable at $$x = 0$$ with $$f'(0) = 3$$, we use the definition of the derivative: $$f'(0) = \lim_{h \to 0}\dfrac{f(0+h) - f(0)}{h} = \lim_{h \to 0}\dfrac{f(h) - 1}{h} = 3$$.

Therefore $$\lim_{h \to 0}\dfrac{1}{h}(f(h) - 1) = 3$$.