For a differentiable function f, the value of `lim_(h to 0) ([f(x+h)]^(2) -[f(x)]^(2))/(2h)` is equal to:

Options

A. `[f'(x)]^(2)`
B. `f(x) f'(x)`
C. `1/2 [f'(x)]^(2)`
D. `1/2[[f'(x)]^(2)-[f(x)]^(2)]`

Answer

To solve the limit problem, we need to evaluate the expression:

\[ \lim_{h \to 0} \frac{[f(x+h)]^2 - [f(x)]^2}{2h} \]

### Step-by-Step Solution:

1. **Recognize the Form**: The expression in the limit resembles the difference of squares, which can be factored. Recall that \(a^2 - b^2 = (a - b)(a + b)\). Here, let \(a = f(x+h)\) and \(b = f(x)\).

2. **Apply the Difference of Squares**:
\[ [f(x+h)]^2 - [f(x)]^2 = (f(x+h) - f(x))(f(x+h) + f(x)) \]
Thus, we can rewrite the limit as:
\[ \lim_{h \to 0} \frac{(f(x+h) - f(x))(f(x+h) + f(x))}{2h} \]

3. **Separate the Limit**: We can separate the limit into two parts:
\[ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \cdot \lim_{h \to 0} \frac{f(x+h) + f(x)}{2} \]

4. **Evaluate the First Limit**: The first limit is the definition of the derivative of \(f\) at \(x\):
\[ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f'(x) \]

5. **Evaluate the Second Limit**: As \(h\) approaches 0, \(f(x+h)\) approaches \(f(x)\):
\[ \lim_{h \to 0} \frac{f(x+h) + f(x)}{2} = \frac{f(x) + f(x)}{2} = f(x) \]

6. **Combine the Results**: Now we can combine the results from the two limits:
\[ \lim_{h \to 0} \frac{[f(x+h)]^2 - [f(x)]^2}{2h} = f'(x) \cdot f(x) \]

### Final Answer:
Thus, the value of the limit is:
\[ \frac{f(x) \cdot f'(x)}{2} \]