Let `f(x) = 1 /(sqrt( 18 - x^2)` The value of `Lt_(x -> 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 \( \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3} \) where \( f(x) = \frac{1}{\sqrt{18 - x^2}} \), we can follow these steps:

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

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

### Step 3: Check for indeterminate form
If we substitute \( x = 3 \) directly into the limit, we get:
\[ f(3) - f(3) = 0 \quad \text{and} \quad x - 3 = 0 \]
This results in the indeterminate form \( \frac{0}{0} \).

### Step 4: Apply L'Hôpital's Rule
Since we have the indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator:
\[ \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3} = \lim_{x \to 3} \frac{f'(x)}{1} \]

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

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

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

### Final Answer
The value of \( \lim_{x \to 3} \frac{f(x) - f(3)}{x - 3} \) is \( \frac{1}{9} \).

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