`Lt_(x to 0) f(1- cos 3x)/x^(2)` where `f(x)` is a continous function satisfying the condition `f(9/2) =2/9` is equal to:

Options

A. 0
B. `9//2`
C. `2//9`
D. none

Answer

To solve the limit \( \lim_{x \to 0} \frac{f(1 - \cos 3x)}{x^2} \) where \( f(x) \) is a continuous function satisfying \( f\left(\frac{9}{2}\right) = \frac{2}{9} \), we can follow these steps:

### Step 1: Simplify the expression inside the limit
We know from trigonometric identities that:
\[ 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \]
Applying this to \( \cos 3x \):
\[ 1 - \cos 3x = 2 \sin^2\left(\frac{3x}{2}\right) \]
Thus, we can rewrite the limit:
\[ \lim_{x \to 0} \frac{f(1 - \cos 3x)}{x^2} = \lim_{x \to 0} \frac{f\left(2 \sin^2\left(\frac{3x}{2}\right)\right)}{x^2} \]

### Step 2: Substitute \( 2 \sin^2\left(\frac{3x}{2}\right) \)
Now, we will express \( \sin^2\left(\frac{3x}{2}\right) \) in terms of \( x \):
\[ \sin\left(\frac{3x}{2}\right) \approx \frac{3x}{2} \quad \text{as } x \to 0 \]
Thus:
\[ \sin^2\left(\frac{3x}{2}\right) \approx \left(\frac{3x}{2}\right)^2 = \frac{9x^2}{4} \]
So we have:
\[ 2 \sin^2\left(\frac{3x}{2}\right) \approx 2 \cdot \frac{9x^2}{4} = \frac{9x^2}{2} \]

### Step 3: Substitute back into the limit
Now we substitute this back into our limit:
\[ \lim_{x \to 0} \frac{f\left(\frac{9x^2}{2}\right)}{x^2} \]
This can be simplified to:
\[ \lim_{x \to 0} \frac{f\left(\frac{9}{2} \cdot x^2\right)}{x^2} \]
Let \( y = \frac{9}{2} x^2 \). As \( x \to 0 \), \( y \to 0 \) as well. Therefore, we can rewrite the limit in terms of \( y \):
\[ \lim_{y \to 0} \frac{f(y)}{\frac{2y}{9}} = \lim_{y \to 0} \frac{9}{2} f(y) \]

### Step 4: Evaluate the limit using continuity
Since \( f \) is continuous and we know \( f\left(\frac{9}{2}\right) = \frac{2}{9} \):
\[ \lim_{y \to 0} f(y) = f(0) \]
Thus, we need to find \( f(0) \). However, we do not have the value of \( f(0) \) directly, but we can evaluate:
\[ \lim_{y \to 0} \frac{9}{2} f(y) = \frac{9}{2} \cdot f(0) \]

### Step 5: Conclusion
Since \( f\left(\frac{9}{2}\right) = \frac{2}{9} \), we can conclude:
\[ \lim_{x \to 0} \frac{f(1 - \cos 3x)}{x^2} = \frac{9}{2} \cdot \frac{2}{9} = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]