If f(3)=6 and f'(3)=2, then `lim_(x to 3) (xf (3)-3f(x))/(x-3)` is given by

Options

A. 6
B. 4
C. 0
D. none of these

Explanation

To solve the limit problem given by:

\[ \lim_{x \to 3} \frac{x f(3) - 3 f(x)}{x - 3} \]

we will follow these steps:

### Step 1: Substitute the known values
We know that \( f(3) = 6 \). So we can substitute this value into the limit expression:

\[ \lim_{x \to 3} \frac{x \cdot 6 - 3 f(x)}{x - 3} \]

### Step 2: Simplify the expression
This simplifies to:

\[ \lim_{x \to 3} \frac{6x - 3f(x)}{x - 3} \]

### Step 3: Factor out the expression
We can rewrite the numerator:

\[ 6x - 3f(x) = 3(2x - f(x)) \]

Thus, our limit becomes:

\[ \lim_{x \to 3} \frac{3(2x - f(x))}{x - 3} \]

### Step 4: Apply L'Hôpital's Rule
As \( x \to 3 \), both the numerator and denominator approach 0, leading to an 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:

\[ \lim_{x \to 3} \frac{3(2 - f'(x))}{1} \]

### Step 5: Evaluate the limit
Now we can substitute \( x = 3 \) into the expression. We know \( f'(3) = 2 \):

\[ = 3(2 - f'(3)) = 3(2 - 2) = 3 \cdot 0 = 0 \]

### Final Result
Thus, the limit is:

\[ \lim_{x \to 3} \frac{x f(3) - 3 f(x)}{x - 3} = 0 \]