Let f: R rightarrow R satisfy the equation f( | vedclass

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(A) Given the functional equation $f(x+y)=f(x) \cdot f(y)$.By definition of the derivative at $x=0$,we have $f'(0) = \lim_{h \rightarrow 0} \frac{f(0+

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$3$

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(A) Given the functional equation $f(x+y)=f(x) \cdot f(y)$.By definition of the derivative at $x=0$,we have $f'(0) = \lim_{h ightarrow 0} \frac{f(0+h)-f(0)}{h}$.Since $f(x+y)=f(x)f(y)$,setting $x=0, y=0$ gives $f(0)=f(0)^2$. Since $f(x) 
eq 0$,we must have $f(0)=1$.Substituting $f(0)=1$ into the derivative definition,we get $f'(0) = \lim_{h ightarrow 0} \frac{f(h)-1}{h}$.Given $f'(0)=3$,it follows that $\lim_{h ightarrow 0} \frac{f(h)-1}{h} = 3$.

Let $f: R \rightarrow R$ 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 $\lim_{h \rightarrow 0} \frac{1}{h}(f(h)-1)$ is equal to ....... .

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