If f (2) = 4 and `f^,(2) =1` then `lim_(x rarr 2) (x f(2)-2 f(x))/(x -2) ` is equal to

Options

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

Answer

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

**Step 1: Identify the limit expression.**
We need to evaluate:
\[ \lim_{x \to 2} \frac{x f(2) - 2 f(x)}{x - 2} \]
Given that \( f(2) = 4 \) and \( f'(2) = 1 \).

**Step 2: Substitute the known value of \( f(2) \).**
Substituting \( f(2) = 4 \) into the limit expression, we have:
\[ \lim_{x \to 2} \frac{x \cdot 4 - 2 f(x)}{x - 2} = \lim_{x \to 2} \frac{4x - 2f(x)}{x - 2} \]

**Step 3: Evaluate the limit directly.**
If we substitute \( x = 2 \) directly, 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 4: 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 can rewrite the limit as:
\[ \lim_{x \to 2} \frac{4 - 2f'(x)}{1} \]

**Step 5: Substitute \( x = 2 \) again.**
Now substituting \( x = 2 \) into the limit expression:
\[ 4 - 2f'(2) = 4 - 2(1) = 4 - 2 = 2 \]

**Final Answer:**
Thus, the limit is:
\[ \lim_{x \to 2} \frac{x f(2) - 2 f(x)}{x - 2} = 2 \]