Question:

Step 1: Rewrite the limit by adding and subtracting $f(3)$ in the numerator. We start with the given limit: $$\lim_{h \to 0} \frac{f(3+7h) - f(3+4h)}{h}$$ To connect this to the derivative, we add and subtract $f(3)$ in the numerator: $$\lim_{h \to 0} \frac{f(3+7h) - f(3) - (f(3+4h) - f(3))}{h}$$ Step 2: Separate the limit into two parts using the linearity of limits. We can split this into two separate limits: $$\lim_{h \to 0} \frac{f(3+7h) - f(3)}{h} - \lim_{h \to 0} \frac{f(3+4h) - f(3)}{h}$$ Step 3: Apply the definition of the derivative to each limit. Recall that the derivative of $f$ at $x = 3$ is defined as: $$f'(3) = \lim_{h \to 0} \frac{f(3+h) - f(3)}{h}$$ For the first limit, we have $f(3+7h)$ in the numerator. We can rewrite this as: $$\lim_{h \to 0} \frac{f(3+7h) - f(3)}{h} = \lim_{h \to 0} \frac{f(3+7h) - f(3)}{7h} \cdot 7 = 7 \cdot f'(3)$$ For the second limit, we have $f(3+4h)$ in the numerator. Similarly: $$\lim_{h \to 0} \frac{f(3+4h) - f(3)}{h} = \lim_{h \to 0} \frac{f(3+4h) - f(3)}{4h} \cdot 4 = 4 \cdot f'(3)$$ Step 4: Combine the results and solve for $f'(3)$. Substituting back into our expression: $$7 \cdot f'(3) - 4 \cdot f'(3) = 4$$ $$3 \cdot f'(3) = 4$$ $$f'(3) = \frac{4}{3}$$ **Final Answer:** The value of $f'(3)$ equals $\dfrac{4}{3}$. The correct option is **Option 3**.