If the normal to `y=f(x)` at (0, 0) is given by `y-x =0,` then `lim_(x to 0)(x^(2))/(f(x^(2))-20f(9x^(2))+2f(99x^(2)))`, is

Options

A. `1//19`
B. `-1//19`
C. `1//2`
D. does not exist

Answer

To solve the problem, we will follow a systematic approach step by step.

### Step 1: Understand the given information
We are given that the normal to the curve \( y = f(x) \) at the point \( (0, 0) \) is represented by the equation \( y - x = 0 \). This implies that the slope of the normal line is 1. Therefore, the slope of the tangent line at this point must be the negative reciprocal, which is -1.

### Step 2: Find the derivative at the point
From the information above, we can conclude:
\[ f'(0) = -1 \]

### Step 3: Evaluate the limit expression
We need to evaluate the limit:
\[ \lim_{x \to 0} \frac{x^2}{f(x^2) - 20f(9x^2) + 2f(99x^2)} \]
As \( x \to 0 \), \( f(x^2) \), \( f(9x^2) \), and \( f(99x^2) \) all approach \( f(0) \), which is 0. Thus, we have a \( \frac{0}{0} \) indeterminate form.

### Step 4: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to a} \frac{h(x)}{g(x)} = \lim_{x \to a} \frac{h'(x)}{g'(x)} \]
if the limit on the right exists.

Let:
- \( h(x) = x^2 \)
- \( g(x) = f(x^2) - 20f(9x^2) + 2f(99x^2) \)

Now we differentiate \( h(x) \) and \( g(x) \):
\[ h'(x) = 2x \]

For \( g'(x) \), we use the chain rule:
\[ g'(x) = f'(x^2) \cdot 2x - 20 \cdot f'(9x^2) \cdot 18x - 2 \cdot f'(99x^2) \cdot 198x \]
Thus,
\[ g'(x) = 2x f'(x^2) - 360x f'(9x^2) - 396x f'(99x^2) \]

### Step 5: Substitute into the limit
Now we substitute back into the limit:
\[ \lim_{x \to 0} \frac{2x}{2x f'(x^2) - 360x f'(9x^2) - 396x f'(99x^2)} \]
We can simplify this to:
\[ \lim_{x \to 0} \frac{2}{2f'(x^2) - 360f'(9x^2) - 396f'(99x^2)} \]

### Step 6: Evaluate the derivatives at \( x = 0 \)
As \( x \to 0 \):
- \( f'(x^2) \to f'(0) = -1 \)
- \( f'(9x^2) \to f'(0) = -1 \)
- \( f'(99x^2) \to f'(0) = -1 \)

Substituting these values:
\[ \lim_{x \to 0} \frac{2}{2(-1) - 360(-1) - 396(-1)} = \frac{2}{-2 + 360 + 396} \]
Calculating the denominator:
\[ -2 + 360 + 396 = 754 \]
Thus, we have:
\[ \lim_{x \to 0} \frac{2}{754} = \frac{1}{377} \]

### Final Answer
The limit evaluates to:
\[ \frac{1}{377} \]