If `f(2) = 4` and `f'(2) =4`, then `"lt"_(x to 2)(xf(2) -2f(x))/(x-2)` =

Options

A. `-4`
B. `-2`
C. 2
D. 3

Answer

To solve the limit problem given by

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

we can follow these steps:

### Step 1: Substitute the known values
We know that \( f(2) = 4 \). Substituting this into the limit gives us:

\[ \lim_{x \to 2} \frac{x \cdot 4 - 2f(x)}{x - 2} = \lim_{x \to 2} \frac{4x - 2f(x)}{x - 2} \]

### Step 2: Evaluate the limit
Now, if we directly substitute \( x = 2 \) into the expression, we get:

\[ \frac{4(2) - 2f(2)}{2 - 2} = \frac{8 - 8}{0} = \frac{0}{0} \]

This is an indeterminate form, so we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if we have an indeterminate form \( \frac{0}{0} \), we can take the derivatives of the numerator and the denominator:

1. **Numerator**: The derivative of \( 4x - 2f(x) \) is \( 4 - 2f'(x) \).
2. **Denominator**: The derivative of \( x - 2 \) is \( 1 \).

Now we can rewrite the limit:

\[ \lim_{x \to 2} \frac{4 - 2f'(x)}{1} \]

### Step 4: Substitute \( x = 2 \) again
Now we substitute \( x = 2 \):

\[ 4 - 2f'(2) \]

We know from the problem statement that \( f'(2) = 4 \). Therefore:

\[ 4 - 2(4) = 4 - 8 = -4 \]

### Final Answer
Thus, the limit is:

\[ \boxed{-4} \]