Question:
Let $f: \mathrm{R} \rightarrow \mathrm{R}$ 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 $\mathrm{x}=0$ and $f^{\prime}(0)=3$, then $\lim _{h \rightarrow 0} \frac{1}{h}(f(h)-1)$ is equal to_________.
Solution:
If $f(x+y)=f(x) \cdot f(y) \& f^{\prime}(0)=3$ then
$f(x)=a^{x} \Rightarrow f^{\prime}(x)=a^{x} \cdot \ell n a$
$\Rightarrow f^{\prime}(0)=\ell \mathrm{na}=3 \Rightarrow \mathrm{a}=\mathrm{e}^{3}$
$\Rightarrow f(x)=\left(e^{3}\right)^{x}=e^{3 x}$
$\lim _{x \rightarrow 0} \frac{f(x)-1}{x}=\lim _{x \rightarrow 0}\left(\frac{\mathrm{e}^{3 x}-1}{3 x} \times 3\right)=1 \times 3=3$