Let `f: R rarr R` be a function such that `f(2)=4 and f'(2)=1`. Then, the value of `lim_(x rarr 2) (x^(2)f(2)-4f(x))/(x-2)` is equal to

Options

A. 4
B. 8
C. 16
D. 12

Answer

To solve the limit problem given in the question, we will follow these steps:

### Step 1: Substitute the values into the limit expression
We start with the limit expression:
\[ \lim_{x \to 2} \frac{x^2 f(2) - 4f(x)}{x - 2} \]
Substituting \( f(2) = 4 \):
\[ \lim_{x \to 2} \frac{x^2 \cdot 4 - 4f(x)}{x - 2} \]
This simplifies to:
\[ \lim_{x \to 2} \frac{4x^2 - 4f(x)}{x - 2} \]

### Step 2: Evaluate the limit directly
Now, we can substitute \( x = 2 \) directly into the expression:
\[ 4(2^2) - 4f(2) = 4 \cdot 4 - 4 \cdot 4 = 16 - 16 = 0 \]
Since both the numerator and denominator approach 0, we have an indeterminate form \( \frac{0}{0} \).

### Step 3: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that if the limit results in \( \frac{0}{0} \), we can differentiate the numerator and the denominator:
\[ \lim_{x \to 2} \frac{d}{dx}(4x^2 - 4f(x)) \Big/ \frac{d}{dx}(x - 2) \]
Calculating the derivatives:
- The derivative of the numerator \( 4x^2 - 4f(x) \) is:
\[ 8x - 4f'(x) \]
- The derivative of the denominator \( x - 2 \) is:
\[ 1 \]

### Step 4: Substitute \( x = 2 \) again
Now we substitute \( x = 2 \) into the differentiated limit:
\[ \lim_{x \to 2} \frac{8x - 4f'(x)}{1} = 8(2) - 4f'(2) \]
Substituting \( f'(2) = 1 \):
\[ = 16 - 4 \cdot 1 = 16 - 4 = 12 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{12} \]