If `f'(x_(0))` exists, then `lim_(h to 0)([f(x_(0) +h) -f(x_(0) -h))/(2h))` is equal to:

Options

A. `1/2f'(x_(0))`
B. `f'(x_(0))`
C. `2f'(x_(0))`
D. none of these

Answer

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

\[ \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0 - h)}{2h} \]

Given that \( f'(x_0) \) exists, we can use the definition of the derivative and properties of limits.

### Step 1: Rewrite the expression

We can express the limit as:

\[ \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0 - h)}{2h} \]

### Step 2: Split the limit

Using the definition of the derivative, we can express \( f'(x_0) \) as:

\[ f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} \]

And similarly, we can write:

\[ f'(x_0) = \lim_{h \to 0} \frac{f(x_0) - f(x_0 - h)}{h} \]

### Step 3: Combine the derivatives

Now, we can rewrite the limit we want to evaluate:

\[ \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0 - h)}{2h} = \frac{1}{2} \left( \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} + \lim_{h \to 0} \frac{f(x_0) - f(x_0 - h)}{h} \right) \]

### Step 4: Substitute the derivatives

Since both limits on the right-hand side are equal to \( f'(x_0) \), we can substitute:

\[ \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} = f'(x_0) \]
\[ \lim_{h \to 0} \frac{f(x_0) - f(x_0 - h)}{h} = f'(x_0) \]

### Step 5: Final evaluation

Thus, we have:

\[ \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0 - h)}{2h} = \frac{1}{2} \left( f'(x_0) + f'(x_0) \right) = \frac{1}{2} \cdot 2f'(x_0) = f'(x_0) \]

### Conclusion

Therefore, the final result is:

\[ \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0 - h)}{2h} = f'(x_0) \]