If f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2 then \(\mathop {\lim }\limits_{x \to a} .\frac{{g(x)f(a) - g(a)f(x)}}{{x - a}}\) is

Given:

f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2

Concept:

L'Hospital's Rule: It tells us that if we have an indeterminate form '0/0' or '∞/∞' all we need to do is differentiate the numerator and the denominator separately and then take the limit.

Calculation: 

\(Let\ \ y\ = \ \mathop {\lim }\limits_{x \to a} .\frac{{g(x)f(a) - g(a)f(x)}}{{x - a}}\)

when x = a, it gives 0/0 form. Hence, we can solve the limit by simply using L'Hospital's rule. 

Hence, differentiate the numerator and the denominator separately with respect to x, we will get

\(⇒ \ y\ = \ \mathop {\lim }\limits_{x \to a} .\frac{{g'(x)f(a) - g(a)f'(x)}}{{ 1}}\)

Put x = a

 \(⇒ \ y\ = \ \mathop {{g'(a)f(a) - g(a)f'(a)}}\)

But according to question,

f(a) = 2, f'(a) = 1, g(a) = -1, g'(a) = 2

⇒ y = (2)(2) - (-1)(1) 

⇒ y = 5

Hence, the value of above limit is 5.