If the normal of `f(x)=0` at x=0 is given by `3x-y+3=0`, then `lim_(x to 0) x^(2)/(f(x^(2)) -5f[4x^(2)) + 4f(7x^(2)))` is equal to:

Options

A. `-1/3`
B. `1/3`
C. `1/6`
D. 0

Answer

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

\[ \lim_{x \to 0} \frac{x^2}{f(x^2) - 5f(4x^2) + 4f(7x^2)} \]

Given that the normal to the function \( f(x) \) at \( x = 0 \) is described by the equation \( 3x - y + 3 = 0 \), we can rewrite this in slope-intercept form:

\[ y = 3x + 3 \]

From this, we can identify the slope of the normal line as \( 3 \). The slope of the tangent line at this point is the negative reciprocal of the slope of the normal. Therefore, we have:

\[ f'(0) = -\frac{1}{3} \]

Next, we need to evaluate the limit expression. First, we substitute \( x = 0 \) into the limit expression:

\[ f(0) - 5f(0) + 4f(0) = 0 \]

This results in an indeterminate form \( \frac{0}{0} \), which allows us to apply L'Hôpital's Rule. According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately.

1. **Differentiate the numerator**:
\[ \frac{d}{dx}(x^2) = 2x \]

2. **Differentiate the denominator**:
We need to differentiate \( f(x^2) - 5f(4x^2) + 4f(7x^2) \):
- For \( f(x^2) \), using the chain rule:
\[ \frac{d}{dx}(f(x^2)) = f'(x^2) \cdot 2x \]
- For \( -5f(4x^2) \):
\[ \frac{d}{dx}(-5f(4x^2)) = -5f'(4x^2) \cdot 8x = -40xf'(4x^2) \]
- For \( 4f(7x^2) \):
\[ \frac{d}{dx}(4f(7x^2)) = 4f'(7x^2) \cdot 14x = 56xf'(7x^2) \]

Combining these, we have:

\[ \frac{d}{dx}(f(x^2) - 5f(4x^2) + 4f(7x^2)) = 2xf'(x^2) - 40xf'(4x^2) + 56xf'(7x^2) \]

Now substituting these derivatives back into the limit gives us:

\[ \lim_{x \to 0} \frac{2x}{2xf'(x^2) - 40xf'(4x^2) + 56xf'(7x^2)} \]

We can cancel \( x \) from the numerator and denominator (since we are approaching \( x = 0 \), \( x \neq 0 \)):

\[ \lim_{x \to 0} \frac{2}{2f'(x^2) - 40f'(4x^2) + 56f'(7x^2)} \]

Now, substituting \( x = 0 \):

\[ = \frac{2}{2f'(0) - 40f'(0) + 56f'(0)} = \frac{2}{(2 - 40 + 56)f'(0)} = \frac{2}{18f'(0)} \]

Since we know \( f'(0) = -\frac{1}{3} \):

\[ = \frac{2}{18 \cdot -\frac{1}{3}} = \frac{2}{-6} = -\frac{1}{3} \]

Thus, the limit evaluates to:

\[ \lim_{x \to 0} \frac{x^2}{f(x^2) - 5f(4x^2) + 4f(7x^2)} = -\frac{1}{3} \]

### Final Answer:
\[ \text{The limit is } -\frac{1}{3}. \]