Let f : R o R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of \mathop {\lim }\limits_{x o 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}} is equal to :

Question

Let f : R 	o R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of \mathop {\lim }\limits_{x 	o 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}} is equal to :

Accepted Answer

Answer: D. 12 This limit can be solved using L'Hopital's Rule, which states that for the limit of the form 0/0 or ±∞/±∞, the limit can be found by taking the derivative of the numerator and the derivative of the denominator separately. \mathop {\lim }\limits_{x 	o 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}} is in the indeterminate form, and we are given that f(2) = 4 and f'(2) = 1, so we can apply L'Hopital's rule. Taking the derivative of the numerator and the denominator, we get : Numerator: derivative of x^2 f(2) - 4f(x) is 2x f(2) - 4f'(x). Denominator: derivative of x - 2 is 1. So, the limit becomes : \mathop {\lim }\limits_{x 	o 2} {{{2xf(2) - 4f'(x)}} \over 1} = 2 	imes 2 	imes f(2) - 4 	imes f'(2) = 16 - 4 = 12. Therefore, Option D, 12, is the correct answer.

Let f : R $$\to$$ R be a function such that f(2) = 4 and f'(2) = 1. Then, the value of $$\mathop {\lim }\limits_{x \to 2} {{{x^2}f(2) - 4f(x)} \over {x - 2}}$$ is equal to :

Answer

Explanation

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