We are asked to find the derivative of the function $f(x) = \sqrt{7 - 4x}$ using the limit definition of the derivative.

The limit definition of the derivative is given by: $f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Let's apply this to $f(x) = \sqrt{7 - 4x}$.

1. First, compute $f(x + h)$: $f(x + h) = \sqrt{7 - 4(x + h)} = \sqrt{7 - 4x - 4h}$

2. Now, substitute this into the limit definition: $f'(x) = \lim_{h \to 0} \frac{\sqrt{7 - 4x - 4h} - \sqrt{7 - 4x}}{h}$

3. To simplify, we rationalize the numerator by multiplying the numerator and denominator by the conjugate: $f'(x) = \lim_{h \to 0} \frac{\left( \sqrt{7 - 4x - 4h} - \sqrt{7 - 4x} \right) \cdot \left( \sqrt{7 - 4x - 4h} + \sqrt{7 - 4x} \right)}{h \cdot \left( \sqrt{7 - 4x - 4h} + \sqrt{7 - 4x} \right)}$

4. The numerator simplifies using the difference of squares: $= \lim_{h \to 0} \frac{(7 - 4x - 4h) - (7 - 4x)}{h \left( \sqrt{7 - 4x - 4h} + \sqrt{7 - 4x} \right)}$ Which simplifies further to: $= \lim_{h \to 0} \frac{-4h}{h \left( \sqrt{7 - 4x - 4h} + \sqrt{7 - 4x} \right)}$

5. Cancel out the $h$ in the numerator and denominator: $= \lim_{h \to 0} \frac{-4}{\sqrt{7 - 4x - 4h} + \sqrt{7 - 4x}}$

6. Finally, take the limit as $h \to 0$: $f'(x) = \frac{-4}{2\sqrt{7 - 4x}} = \frac{-2}{\sqrt{7 - 4x}}$

Thus, the derivative of the function is: $f'(x) = \frac{-2}{\sqrt{7 - 4x}}$

Would you like more details on any step?

Here are some related questions:

1. How can we interpret the result geometrically?
2. Can we use the chain rule to verify this derivative?
3. What happens to $f'(x)$ as $x$ approaches a value where the function becomes undefined?
4. How would we apply this technique to a more complex radical function?
5. How does rationalizing the numerator help simplify the limit?

Tip: Rationalizing expressions with square roots is a key step in simplifying many limit problems.