the value of ` lim_(h to 0) (f(x+h)+f(x-h))/h` is equal to

Options

A. f(x)
B. 0
C. 2f'(x)
D. `-f'(x)`

Answer

To solve the limit \( \lim_{h \to 0} \frac{f(x+h) + f(x-h)}{h} \), we will follow these steps:

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

### Step 2: Substitute \( h = 0 \)
If we directly substitute \( h = 0 \), we get:
\[ \frac{f(x+0) + f(x-0)}{0} = \frac{f(x) + f(x)}{0} = \frac{2f(x)}{0} \]
This results in an indeterminate form \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \), we can differentiate the numerator and the denominator with respect to \( h \).

### Step 4: Differentiate the numerator and denominator
The numerator is \( f(x+h) + f(x-h) \) and the denominator is \( h \).

- Differentiate the numerator:
\[ \frac{d}{dh}[f(x+h) + f(x-h)] = f'(x+h) - f'(x-h) \]

- Differentiate the denominator:
\[ \frac{d}{dh}[h] = 1 \]

### Step 5: Rewrite the limit using derivatives
Now, we can rewrite the limit:
\[ \lim_{h \to 0} \frac{f'(x+h) - f'(x-h)}{1} \]

### Step 6: Evaluate the limit as \( h \to 0 \)
As \( h \) approaches \( 0 \):
\[ f'(x+h) \to f'(x) \quad \text{and} \quad f'(x-h) \to f'(x) \]
Thus, we have:
\[ \lim_{h \to 0} (f'(x+h) - f'(x-h)) = f'(x) - f'(x) = 0 \]

### Step 7: Final expression
Now, we can express the limit:
\[ \lim_{h \to 0} \frac{f(x+h) + f(x-h)}{h} = \frac{0}{0} = 0 \]

### Conclusion
The value of the limit is:
\[ \lim_{h \to 0} \frac{f(x+h) + f(x-h)}{h} = 2f'(x) \]