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Step-by-step Solution
Step 1: Understand the Functional Equation
We are given a function f:\mathbb{R} \to \mathbb{R} satisfying
f(x + y) = f(x)\,f(y) for all real x,y , and additionally f(x) \neq 0 for any x \in \mathbb{R} . Such a functional equation typically implies an exponential form for f(x) .
Step 2: Propose the Exponential Form
From the condition f(x + y) = f(x)\,f(y) , we assume a solution of the form
f(x) = a^x \text{ for some constant }a \neq 0.
Step 3: Use the Given Derivative Information
We know f is differentiable at x = 0 and f'(0) = 3 . First, compute the derivative of f(x) in the proposed form:
f(x) = a^x \quad \Longrightarrow \quad f'(x) = a^x \ln(a).
In particular, at x = 0 ,
f'(0) = a^0 \ln(a) = \ln(a).
Since f'(0) = 3 , we get
\ln(a) = 3 \quad \Longrightarrow \quad a = e^3.
Step 4: Rewrite f(x) and Evaluate the Limit
With a = e^3 , the function becomes
f(x) = (e^3)^x = e^{3x}.
Hence,
f(h) = e^{3h}.
We need to find
\displaystyle \lim_{h \to 0} \frac{f(h) - 1}{h}
= \lim_{h \to 0} \frac{e^{3h} - 1}{h}.
Step 5: Compute the Limit
Recall the standard limit
\displaystyle \lim_{u \to 0} \frac{e^u - 1}{u} = 1.
In our case, let u = 3h , so as h \to 0 , we also have u \to 0 . Then,
\[
\lim_{h \to 0} \frac{e^{3h} - 1}{h}
= \lim_{h \to 0} \Biggl(\frac{e^{3h} - 1}{3h} \times 3\Biggr)
= 3 \times \lim_{h \to 0} \frac{e^{3h} - 1}{3h}
= 3 \times 1
= 3.
\]
Therefore, the value of the limit is \boxed{3}.
