If f ' (3)=1 , then `Lt_(h to0)(f(3+h)-f(3-h))/(h)` is

Options

A. 1
B. 2
C. 4
D. `(1)/(2)`

Answer

To solve the limit \( \lim_{h \to 0} \frac{f(3+h) - f(3-h)}{h} \), given that \( f'(3) = 1 \), we can follow these steps:

### Step 1: Rewrite the Limit
We start with the expression:
\[ \lim_{h \to 0} \frac{f(3+h) - f(3-h)}{h} \]

### Step 2: Apply the Definition of Derivative
We can recognize that the expression resembles the definition of the derivative. Specifically, we can express \( f(3+h) \) and \( f(3-h) \) in terms of \( f'(3) \):
\[ f(3+h) = f(3) + f'(3)h + o(h) \quad \text{(as } h \to 0\text{)} \]
\[ f(3-h) = f(3) - f'(3)h + o(h) \quad \text{(as } h \to 0\text{)} \]

### Step 3: Substitute into the Limit
Substituting these expressions into our limit gives:
\[ \lim_{h \to 0} \frac{(f(3) + f'(3)h + o(h)) - (f(3) - f'(3)h + o(h))}{h} \]

### Step 4: Simplify the Expression
This simplifies to:
\[ \lim_{h \to 0} \frac{f'(3)h + f'(3)h + o(h) - o(h)}{h} = \lim_{h \to 0} \frac{2f'(3)h}{h} \]
Assuming \( h \neq 0 \), we can cancel \( h \):
\[ \lim_{h \to 0} 2f'(3) = 2f'(3) \]

### Step 5: Substitute the Known Value
Since we know \( f'(3) = 1 \):
\[ 2f'(3) = 2 \times 1 = 2 \]

### Conclusion
Thus, the limit evaluates to:
\[ \lim_{h \to 0} \frac{f(3+h) - f(3-h)}{h} = 2 \]

### Final Answer
The value of the limit is \( 2 \).

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