If `f(a) =2, f'(a)=1, g(a) =-1 , g' (a)=2`, then the value of `lim_(xrarr a) (g(x)f(a)-g(a)f(x))/(x-a)`, is

Options

A. `-5`
B. `1//5`
C. 5
D. `-1//5`

Answer

To solve the limit problem given by

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

we can follow these steps:

### Step 1: Substitute the known values
We know that \( f(a) = 2 \), \( g(a) = -1 \), \( f'(a) = 1 \), and \( g'(a) = 2 \). Let's substitute these values into the limit expression.

### Step 2: Evaluate the limit directly
Substituting \( x = a \) directly into the limit gives us:

\[ g(a)f(a) - g(a)f(a) = -1 \cdot 2 - (-1) \cdot 2 = -2 + 2 = 0. \]

The denominator \( x - a \) also approaches \( 0 \) as \( x \to a \). Thus, we have a \( \frac{0}{0} \) indeterminate form.

### Step 3: Apply L'Hôpital's Rule
Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the derivative of the denominator separately.

### Step 4: Differentiate the numerator and denominator
The numerator is \( g(x)f(a) - g(a)f(x) \). Differentiating it with respect to \( x \):

\[ \frac{d}{dx}[g(x)f(a) - g(a)f(x)] = g'(x)f(a) - g(a)f'(x). \]

The denominator \( x - a \) differentiates to \( 1 \).

### Step 5: Substitute \( x = a \) again
Now we substitute \( x = a \) into the derivatives:

\[ \lim_{x \to a} \left( g'(x)f(a) - g(a)f'(x) \right) = g'(a)f(a) - g(a)f'(a). \]

Substituting the known values:

\[ g'(a) = 2, \quad f(a) = 2, \quad g(a) = -1, \quad f'(a) = 1. \]

So we have:

\[ 2 \cdot 2 - (-1) \cdot 1 = 4 + 1 = 5. \]

### Final Answer
Thus, the value of the limit is

\[ \boxed{5}. \]