Let `f (2) = 4 and f '(2) =4. If lim _( xto 2) (xf (2) -2 f (x))/(x-2)=-a,` then a is

Answer

To solve the problem, we need to evaluate the limit:

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

Given that \( f(2) = 4 \) and \( f'(2) = 4 \).

### Step 1: Substitute \( f(2) \) into the limit expression

Substituting \( f(2) = 4 \) into the limit gives:

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

### Step 2: Evaluate the limit directly

Now, substituting \( x = 2 \) directly into the expression results in:

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

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

### Step 3: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, we differentiate the numerator and the denominator:

1. Differentiate the numerator: \( \frac{d}{dx}(4x - 2f(x)) = 4 - 2f'(x) \)
2. Differentiate the denominator: \( \frac{d}{dx}(x - 2) = 1 \)

Thus, we have:

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

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

Now substituting \( x = 2 \):

\[ 4 - 2f'(2) = 4 - 2 \cdot 4 = 4 - 8 = -4 \]

### Step 5: Relate to \( -a \)

From the problem, we know that:

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

Thus, we have:

\[ -4 = -a \implies a = 4 \]

### Final Answer

Therefore, the value of \( a \) is:

\[ \boxed{4} \]