If `f '(2) =1,` then `lim _( h to 0 ) (f (2+ h ^(2)) -f ( 2- h ^(2)))/( 2h ^(2))`is equal to

Options

A. 0
B. 1
C. 2
D. `1//2`

Answer

To solve the limit problem given that \( f'(2) = 1 \), we need to evaluate the following limit:

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

### Step 1: Recognize the form of the limit
As \( h \to 0 \), both \( f(2 + h^2) \) and \( f(2 - h^2) \) approach \( f(2) \). Therefore, the expression becomes:

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

This is an indeterminate form, which suggests that we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Using L'Hôpital's Rule, we differentiate the numerator and denominator with respect to \( h \):

1. **Differentiate the numerator**:
\[ \frac{d}{dh}[f(2 + h^2) - f(2 - h^2)] = f'(2 + h^2) \cdot \frac{d}{dh}(2 + h^2) - f'(2 - h^2) \cdot \frac{d}{dh}(2 - h^2) \]
This simplifies to:
\[ f'(2 + h^2) \cdot 2h - f'(2 - h^2) \cdot (-2h) = 2h[f'(2 + h^2) + f'(2 - h^2)] \]

2. **Differentiate the denominator**:
\[ \frac{d}{dh}[2h^2] = 4h \]

### Step 3: Rewrite the limit
Now we can rewrite the limit as:

\[ \lim_{h \to 0} \frac{2h[f'(2 + h^2) + f'(2 - h^2)]}{4h} \]

### Step 4: Simplify the expression
We can cancel \( h \) from the numerator and denominator (as \( h \to 0 \), \( h \neq 0 \)):

\[ \lim_{h \to 0} \frac{2[f'(2 + h^2) + f'(2 - h^2)]}{4} = \lim_{h \to 0} \frac{f'(2 + h^2) + f'(2 - h^2)}{2} \]

### Step 5: Evaluate the limit
As \( h \to 0 \), both \( 2 + h^2 \) and \( 2 - h^2 \) approach \( 2 \). Thus, we have:

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

Given that \( f'(2) = 1 \):

\[ \lim_{h \to 0} \frac{f(2 + h^2) - f(2 - h^2)}{2h^2} = 1 \]

### Final Answer
Thus, the limit is:

\[ \boxed{1} \]