Key Concepts and Formulas

• Functional Equation: The given equation $f(x + y) = f(x) \cdot f(y)$ is a characteristic functional equation for exponential functions.
• Definition of the Derivative: The derivative of a function $f$ at a point $a$, denoted by $f'(a)$, is defined as: $f'(a) = \mathop {\lim }\limits_{h \to 0} \frac{f(a+h) - f(a)}{h}$
• Standard Limit: The limit $\mathop {\lim }\limits_{k \to 0} \frac{e^k - 1}{k} = 1$ is a fundamental limit in calculus.

Step-by-Step Solution

Step 1: Analyze the given functional equation. We are given the functional equation $f(x + y) = f(x) \cdot f(y)$ for all $x, y \in \mathbb{R}$, and $f(x) \ne 0$. This type of equation implies that $f(x)$ must be of the form $f(x) = a^x$ for some positive constant $a$.

Step 2: Use the differentiability at $x=0$ to find the specific form of $f(x)$. We know that $f$ is differentiable at $x=0$ and $f'(0) = 3$. Let's use the definition of the derivative at $x=0$: $f'(0) = \mathop {\lim }\limits_{h \to 0} \frac{f(0+h) - f(0)}{h}$ First, let's find $f(0)$. Substitute $x=0$ and $y=0$ into the functional equation: $f(0+0) = f(0) \cdot f(0)$ $f(0) = (f(0))^2$ Since $f(x) \ne 0$ for any $x$, $f(0) \ne 0$. Thus, we can divide by $f(0)$: $1 = f(0)$ Now, substitute $f(0)=1$ into the derivative definition: $f'(0) = \mathop {\lim }\limits_{h \to 0} \frac{f(h) - 1}{h}$ We are given that $f'(0) = 3$. Therefore, $3 = \mathop {\lim }\limits_{h \to 0} \frac{f(h) - 1}{h}$

Step 3: Relate the required limit to the definition of the derivative. The expression we need to evaluate is $\mathop {\lim }\limits_{h \to 0} \frac{1}{h}(f(h) - 1)$. This can be rewritten as: $\mathop {\lim }\limits_{h \to 0} \frac{f(h) - 1}{h}$ From Step 2, we have already established that this limit is equal to $f'(0)$.

Step 4: Substitute the given value of $f'(0)$. We are given that $f'(0) = 3$. Therefore, the value of the limit is 3.

Step 5: Verify the result using the assumed form of $f(x)$. As established in Step 1, $f(x) = a^x$. Then, $f'(x) = a^x \ln(a)$. At $x=0$, $f'(0) = a^0 \ln(a) = 1 \cdot \ln(a) = \ln(a)$. Given $f'(0) = 3$, we have $\ln(a) = 3$, which implies $a = e^3$. So, $f(x) = (e^3)^x = e^{3x}$. Now, let's evaluate the limit: $\mathop {\lim }\limits_{h \to 0} \frac{1}{h}(f(h) - 1) = \mathop {\lim }\limits_{h \to 0} \frac{e^{3h} - 1}{h}$ To evaluate this limit, we can use the standard limit $\mathop {\lim }\limits_{k \to 0} \frac{e^k - 1}{k} = 1$. Let $k = 3h$. As $h \to 0$, $k \to 0$. $\mathop {\lim }\limits_{h \to 0} \frac{e^{3h} - 1}{h} = \mathop {\lim }\limits_{h \to 0} \frac{e^{3h} - 1}{3h} \cdot 3$ $= 3 \cdot \mathop {\lim }\limits_{k \to 0} \frac{e^k - 1}{k} = 3 \cdot 1 = 3$ This confirms our result.

Common Mistakes & Tips

• Incorrectly assuming $f(x)=e^x$: While $f(x)=e^x$ satisfies $f(x+y)=f(x)f(y)$, the derivative $f'(0)$ would be 1, not 3. Always use the given $f'(0)$ to determine the base of the exponential.
• Algebraic errors in limit manipulation: Be careful when rewriting the limit expression to match standard forms. Ensure that the substitution (like $k=3h$) is handled correctly.
• Forgetting $f(0)=1$: The functional equation implies $f(0)=1$ if $f(x) \ne 0$. This is crucial for using the definition of the derivative at $x=0$.

Summary

The problem provides a functional equation $f(x+y) = f(x)f(y)$ with $f(x) \ne 0$, implying $f(x) = a^x$. We are also given that $f$ is differentiable at $x=0$ and $f'(0)=3$. The limit to be evaluated, $\mathop {\lim }\limits_{h \to 0} \frac{1}{h}(f(h) - 1)$, is precisely the definition of the derivative of $f$ at $x=0$, i.e., $f'(0)$. Since $f'(0)$ is given as 3, the value of the limit is 3. We verified this by first finding the specific function $f(x) = e^{3x}$ and then evaluating the limit using standard limit properties.

The final answer is \boxed{3}.