Let `f(2)=4` and `f'(2)=4`. Then `lim_(x->2)(xf(2)-2f(x))/(x-2)` is equal to

Options

A. `-1/3`
B. `-2`
C. `-4`
D. 3

Answer

To solve the limit \( \lim_{x \to 2} \frac{xf(2) - 2f(x)}{x - 2} \), given that \( f(2) = 4 \) and \( f'(2) = 4 \), we can follow these steps:

### Step 1: Substitute the known values into the limit expression
We know that \( f(2) = 4 \). Therefore, we can substitute this into the limit expression:

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

This simplifies to:

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

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

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

Since we encounter the indeterminate form \( \frac{0}{0} \), 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:

- The derivative of the numerator \( 4x - 2f(x) \) is \( 4 - 2f'(x) \).
- The derivative of the denominator \( x - 2 \) is \( 1 \).

Thus, we rewrite the limit as:

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

### Step 4: Substitute \( x = 2 \) again
Now we substitute \( x = 2 \) into the new limit expression:

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

Given that \( f'(2) = 4 \):

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

### Final Answer
Thus, the limit is:

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