If the equation of the normal to the curve `y=f(x) at x=0` is `3x-y+3=0` then the value of
`lim_(xrarr0) (x^(2))/({f(x^(2))-5f(4x^(2))+4f(7x^(2))})` is

Options

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

Explanation

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

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

Given that the equation of the normal to the curve \(y = f(x)\) at \(x = 0\) is \(3x - y + 3 = 0\), we can rewrite this as:

\[ y = 3x + 3 \]

From this, we can identify the slope of the normal line, which is \(3\). The slope of the tangent line at \(x = 0\) is the negative reciprocal of the slope of the normal. Therefore, we have:

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

Next, we will use Taylor's expansion to approximate \(f(x^2)\), \(f(4x^2)\), and \(f(7x^2)\) around \(x = 0\):

1. **Taylor Expansion**:
\[ f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + O(x^3) \]

Using this expansion, we can write:

- For \(f(x^2)\):
\[ f(x^2) \approx f(0) + f'(0)x^2 + \frac{f''(0)}{2}x^4 \]

- For \(f(4x^2)\):
\[ f(4x^2) \approx f(0) + f'(0)(4x^2) + \frac{f''(0)}{2}(4x^2)^2 = f(0) + 4f'(0)x^2 + 8f''(0)x^4 \]

- For \(f(7x^2)\):
\[ f(7x^2) \approx f(0) + f'(0)(7x^2) + \frac{f''(0)}{2}(7x^2)^2 = f(0) + 7f'(0)x^2 + \frac{49}{2}f''(0)x^4 \]

2. **Substituting into the limit**:
Now we substitute these approximations into the limit expression:

\[ f(x^2) - 5f(4x^2) + 4f(7x^2) \approx \left(f(0) + f'(0)x^2 + \frac{f''(0)}{2}x^4\right) - 5\left(f(0) + 4f'(0)x^2 + 8f''(0)x^4\right) + 4\left(f(0) + 7f'(0)x^2 + \frac{49}{2}f''(0)x^4\right) \]

Simplifying this gives:

\[ = f(0) + f'(0)x^2 + \frac{f''(0)}{2}x^4 - 5f(0) - 20f'(0)x^2 - 40f''(0)x^4 + 4f(0) + 28f'(0)x^2 + 98f''(0)x^4 \]

Combining like terms:

\[ = (f(0) - 5f(0) + 4f(0)) + (f'(0) - 20f'(0) + 28f'(0))x^2 + \left(\frac{f''(0)}{2} - 40f''(0) + 98f''(0)\right)x^4 \]

This simplifies to:

\[ = 0 + 9f'(0)x^2 + \left(\frac{f''(0)}{2} + 58f''(0)\right)x^4 \]

Thus, we have:

\[ f(x^2) - 5f(4x^2) + 4f(7x^2) \approx 9f'(0)x^2 + O(x^4) \]

3. **Finding the limit**:
Now substituting back into the limit:

\[ \lim_{x \to 0} \frac{x^2}{9f'(0)x^2 + O(x^4)} = \lim_{x \to 0} \frac{1}{9f'(0) + O(x^2)} = \frac{1}{9f'(0)} \]

Substituting \(f'(0) = -\frac{1}{3}\):

\[ \lim_{x \to 0} \frac{1}{9 \cdot -\frac{1}{3}} = \frac{1}{-3} = -\frac{1}{3} \]

Thus, the final answer is:

\[ \boxed{-\frac{1}{3}} \]