Let f''(x) be continuous at x = 0 and f"(0) = 4 then value of `lim_(x->0)(2f(x)-3f(2x)+f(4x))/(x^2)`

Options

A. 11
B. 2
C. 12
D. none of these

Answer

To solve the limit problem, we start with the expression given:

\[ \lim_{x \to 0} \frac{2f(x) - 3f(2x) + f(4x)}{x^2} \]

### Step 1: Check the form of the limit
First, we substitute \( x = 0 \) into the expression:

\[ 2f(0) - 3f(0) + f(0) = 0 \]

This results in \( \frac{0}{0} \) form, which is indeterminate. Therefore, we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

- The numerator:
\[ \frac{d}{dx}(2f(x) - 3f(2x) + f(4x)) = 2f'(x) - 3 \cdot 2f'(2x) + 4f'(4x) = 2f'(x) - 6f'(2x) + 4f'(4x) \]

- The denominator:
\[ \frac{d}{dx}(x^2) = 2x \]

Now we have:

\[ \lim_{x \to 0} \frac{2f'(x) - 6f'(2x) + 4f'(4x)}{2x} \]

### Step 3: Substitute \( x = 0 \) again
Substituting \( x = 0 \) again gives us:

\[ 2f'(0) - 6f'(0) + 4f'(0) = 0 \]

This is again a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 4: Differentiate again
Differentiate the numerator and the denominator again:

- The numerator:
\[ \frac{d}{dx}(2f'(x) - 6f'(2x) + 4f'(4x)) = 2f''(x) - 6 \cdot 2f''(2x) + 4 \cdot 4f''(4x) = 2f''(x) - 12f''(2x) + 16f''(4x) \]

- The denominator:
\[ \frac{d}{dx}(2x) = 2 \]

Now we have:

\[ \lim_{x \to 0} \frac{2f''(x) - 12f''(2x) + 16f''(4x)}{2} \]

### Step 5: Substitute \( x = 0 \) again
Substituting \( x = 0 \):

\[ 2f''(0) - 12f''(0) + 16f''(0) = 2f''(0) - 12f''(0) + 16f''(0) = 6f''(0) \]

Given that \( f''(0) = 4 \):

\[ 6f''(0) = 6 \cdot 4 = 24 \]

### Step 6: Final limit value
Now, divide by 2:

\[ \frac{24}{2} = 12 \]

Thus, the value of the limit is:

\[ \boxed{12} \]