Let `f(x) = 1/(sqrt(18-x^(2)))`. The value of: `lim_(x to 3)(f(x) -f(3))/(x-3)` is:

Options

A. 0
B. `-1//9`
C. `-1//3`
D. `1//9`

Answer

To solve the limit problem, we will follow these steps:

Given the function:
\[ f(x) = \frac{1}{\sqrt{18 - x^2}} \]

We need to find:
\[ \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3} \]

### Step 1: Calculate \( f(3) \)
First, we need to evaluate \( f(3) \):
\[ f(3) = \frac{1}{\sqrt{18 - 3^2}} = \frac{1}{\sqrt{18 - 9}} = \frac{1}{\sqrt{9}} = \frac{1}{3} \]

### Step 2: Substitute into the limit expression
Now substitute \( f(3) \) into the limit expression:
\[ \lim_{x \to 3} \frac{f(x) - \frac{1}{3}}{x - 3} \]

### Step 3: Check the form of the limit
If we directly substitute \( x = 3 \) into the limit, we get:
\[ \frac{f(3) - f(3)}{3 - 3} = \frac{0}{0} \]
This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
\[ \lim_{x \to 3} \frac{f'(x)}{1} \]

### Step 5: Find \( f'(x) \)
To find \( f'(x) \), we use the chain rule:
\[ f(x) = (18 - x^2)^{-\frac{1}{2}} \]
Using the chain rule:
\[ f'(x) = -\frac{1}{2}(18 - x^2)^{-\frac{3}{2}} \cdot (-2x) = \frac{x}{(18 - x^2)^{\frac{3}{2}}} \]

### Step 6: Evaluate \( f'(3) \)
Now we evaluate \( f'(3) \):
\[ f'(3) = \frac{3}{(18 - 3^2)^{\frac{3}{2}}} = \frac{3}{(18 - 9)^{\frac{3}{2}}} = \frac{3}{9^{\frac{3}{2}}} = \frac{3}{27} = \frac{1}{9} \]

### Step 7: Conclusion
Thus, the limit is:
\[ \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3} = f'(3) = \frac{1}{9} \]

So, the final answer is:
\[ \boxed{\frac{1}{9}} \]