`f(x)` is differentiable function at `x=a` such that `f'(a)=2 , f(a)=4.` Find `lim_(xrarra) (xf(a)-af(x))/(x-a)`

Answer

To solve the limit problem, we start with the expression given:

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

### Step 1: Substitute known values
We know that \( f(a) = 4 \) and we can substitute this into the limit:

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

This simplifies to:

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

### Step 2: Analyze the limit
As \( x \) approaches \( a \), both the numerator and denominator approach 0, which gives us an indeterminate form \( \frac{0}{0} \). Therefore, we can apply L'Hôpital's Rule, which states that if we have an indeterminate form, we can take the derivative of the numerator and the denominator.

### Step 3: Differentiate the numerator and denominator
Differentiate the numerator \( 4x - a f(x) \):

- The derivative of \( 4x \) is \( 4 \).
- The derivative of \( -a f(x) \) is \( -a f'(x) \).

So, the derivative of the numerator is:

\[ 4 - a f'(x) \]

Differentiate the denominator \( x - a \):

- The derivative of \( x \) is \( 1 \).
- The derivative of \( -a \) is \( 0 \).

So, the derivative of the denominator is:

\[ 1 \]

### Step 4: Apply L'Hôpital's Rule
Now we can apply L'Hôpital's Rule:

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

### Step 5: Substitute \( x = a \)
Now, substitute \( x = a \) into the limit:

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

We know from the problem statement that \( f'(a) = 2 \). Therefore, we substitute this value:

\[ 4 - a \cdot 2 = 4 - 2a \]

### Final Answer
Thus, the limit is:

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

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