Correct Answer - Option 3 : 5
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.
